Question 14 Marks
Find the cube of the following binomial expressions:$\frac{3}{\text{x}}-\frac{2}{\text{x}^2}$
AnswerGiven,$\Big(\frac{3}{\text{x}}-\frac{2}{\text{x}^2}\Big)^3$
The above equation is in the form of $(a - b)^3= a^3 - b^3 - 3ab(a - b)$
We know that, $\text{a}=\frac{3}{\text{x}},\text{b}=\frac{2}{\text{x}^2}$
By using $(a - b)^3$ formula
$\Big(\frac{3}{\text{x}}-\frac{2}{\text{x}^2}\Big)^3$
$=(3\text{x})^3-\Big(\frac{2}{\text{x}^2}\Big)^3-3\Big(\frac{3}{\text{x}}\Big)\Big(\frac{2}{\text{x}^2}\Big)\Big(\frac{3}{\text{x}}-\frac{2}{\text{x}^2}\Big)$
$=\frac{27}{\text{x}^3}-\frac{8}{\text{x}^6}-3\times\frac{3}{\text{x}}\times\frac{2}{\text{x}^2}\Big(\frac{3}{\text{x}}-\frac{2}{\text{x}^2}\Big)$
$=\frac{27}{\text{x}^3}-\frac{8}{\text{x}^6}-\frac{18}{\text{x}^3}\Big(\frac{3}{\text{x}}-\frac{2}{\text{x}^2}\Big)$
$=\frac{27}{\text{x}^3}-\frac{8}{\text{x}^6}-\Big(\frac{18}{\text{x}^3}\times\frac{3}{\text{x}}\Big)+\Big(\frac{18}{\text{x}^3}\times\frac{2}{\text{x}^2}\Big)$
$=\frac{27}{\text{x}^3}-\frac{8}{\text{x}^6}-\frac{54}{\text{x}^4}+\frac{36}{\text{x}^5}$
Hence $\Big(\frac{3}{\text{x}}-\frac{2}{\text{x}^2}\Big)^3=\frac{27}{\text{x}^3}-\frac{8}{\text{x}^6}-\frac{54}{\text{x}^4}+\frac{36}{\text{x}^5}$
View full question & answer→Question 24 Marks
If $\text{x}^2+\frac{1}{\text{x}^2}=98,$ find the value of $\text{x}^3+\frac{1}{\text{x}^3}.$
AnswerGiven, $\text{x}^2+\frac{1}{\text{x}^2}=98$ We know that, $(x + y)^2 = x^2 + y^2 + 2xy$ ...(1)
Substitute $\text{x}^2+\frac{1}{\text{x}^2}=98$ in eq.(1)
$\Big(\text{x}+\frac{1}{\text{x}}\Big)^2=\text{x}^2+\frac{1}{\text{x}^2}+2\times\text{x}\times\frac{1}{\text{x}}$
$\Big(\text{x}+\frac{1}{\text{x}}\Big)^2=\text{x}^2+\frac{1}{\text{x}^2}+2$
$\Big(\text{x}+\frac{1}{\text{x}}\Big)^2=98+2$
$\Big(\text{x}+\frac{1}{\text{x}}\Big)^2=100$
$\Big(\text{x}+\frac{1}{\text{x}}\Big)^2=\sqrt{100}$
$\Big(\text{x}+\frac{1}{\text{x}}\Big)^2=\pm10$
We need to find $\text{x}^3+\frac{1}{\text{x}^3}$ So,$ a^3 + b^3 = (a - b)(a^2 + b^2 - ab)$$\text{x}^3+\frac{1}{\text{x}^3}=\Big(\text{x}+\frac{1}{\text{x}}\Big)\Big(\text{x}^2+\frac{1}{\text{x}^2}\Big)-\Big(\text{x}\times\frac{1}{\text{x}}\Big)$
We know that,$\Big(\text{x}+\frac{1}{\text{x}}\Big)=10$ and $\Big(\text{x}^2+\frac{1}{\text{x}^2}\Big)=98$
$\Rightarrow\text{x}^3+\frac{1}{\text{x}^3}=10(98-1)$
$\Rightarrow\text{x}^3+\frac{1}{\text{x}^3}=10(97)$
$\Rightarrow\text{x}^3+\frac{1}{\text{x}^3}=970$
Hence, the value of $\text{x}^3+\frac{1}{\text{x}^3}=970.$
View full question & answer→Question 34 Marks
Simplify:
$(a + b + c)^2 + (a - b + c)^2$
AnswerIn the given problem,
we have to simplify the expressions
Given $(a + b + c)^2 + (a - b + c)^2$ By using identity $(x + y + z)^2 = x^2 + y^2 + z^2 + 2xy + 2yz + 2zx$
Hence the equation becomes
$\big(\text{a}+\text{b}+\text{c}\big)+\big(\text{a}\text{b}+\text{c}\big)\\=\Big[\text{a}^2+\text{b}^2+\text{c}^2+2\text{ab}+2\text{bc}+2\text{ca}\Big]\\+\Big[\text{a}^2+(-\text{b})^2+\text{c}^2+2\text{a}(-\text{b})+2(-\text{b})(\text{c})+2\text{ca}\Big]$
$=\text{a}^2+\text{b}^2+\text{c}^2+2\text{ab}+2\text{bc}+2\text{ca}\\+\text{a}^2+\text{b}^2+\text{c}^2-2\text{ab}-2\text{bc}+2\text{ca}$
$=\text{a}^2+\text{a}^2+\text{b}^2+\text{b}^2+\text{c}^2+\text{c}^2\\+2\text{ab}-2\text{ab}+2\text{bc}-2\text{bc}+2\text{ca}+2\text{ca}$
$=2\text{a}^2+2\text{b}^2+2\text{c}^2+4\text{ca}$
Talking 2 as common factor we get$=2\big(\text{a}^2+\text{b}^2+\text{c}^2+2\text{ca}\big)$
Hence the simplified value of $\big(\text{a}+\text{b}+\text{c}\big)^2+\big(\text{a}-\text{b}+\text{c}\big)^2$ is $2\big(\text{a}^2+\text{b}^2+\text{c}^2+2\text{ca}\big).$
View full question & answer→Question 44 Marks
Simplify the following:$\Big(\frac{\text{x}}{2}+\frac{\text{y}}{3}\Big)^3-\Big(\frac{\text{x}}{2}-\frac{\text{y}}{3}\Big)^3$
AnswerGiven $\Big(\frac{\text{x}}{2}+\frac{\text{y}}{3}\Big)^3-\Big(\frac{\text{x}}{2}-\frac{\text{y}}{3}\Big)^3$
We shall use the identity $a^3 - b^3 = (a - b)(a^2 + b^2+ ab)$
Here $\text{a}=\Big(\frac{\text{x}}{2}+\frac{\text{y}}{3}\Big),\ \text{b=}\Big(\frac{\text{x}}{2}-\frac{\text{y}}{3}\Big)$
By applying identity we get$=\bigg(\Big(\frac{\text{x}}{2}+\frac{\text{y}}{3}\Big)-\Big(\frac{\text{x}}{2}-\frac{\text{y}}{3}\Big)\bigg)\\\bigg[\Big(\frac{\text{x}}{2}+\frac{\text{y}}{3}\Big)^2+\Big(\frac{\text{x}}{2}-\frac{\text{y}}{3}\Big)^2-\Big(\frac{\text{x}}{2}+\frac{\text{y}}{3}\Big)\Big(\frac{\text{x}}{2}-\frac{\text{y}}{3}\Big)\bigg]$
$=\Big(\frac{\text{x}}{2}+\frac{\text{y}}{3}-\frac{\text{x}}{2}+\frac{\text{y}}{3}\Big)\Bigg[\bigg(\Big(\frac{\text{x}}{2}\Big)^2+\Big(\frac{\text{y}}{3}\Big)^2+\frac{2\text{xy}}{6}\Big)\\+\bigg(\Big(\frac{\text{x}}{2}\Big)^2+\Big(\frac{\text{y}}{3}\Big)^2-\frac{2\text{xy}}{6}\bigg)+\bigg(\Big(\frac{\text{x}}{2}\Big)^2-\Big(\frac{\text{y}}{3}\Big)^2\bigg)\Bigg]$
$=\frac{2\text{y}}{3}\Big[\Big(\frac{\text{x}^2}{4}+\frac{\text{y}^2}{9}+\frac{2\text{xy}}{6}\Big)+\Big(\frac{\text{x}^2}{4}+\frac{\text{y}^2}{9}-\frac{2\text{xy}}{6}\Big)+\frac{\text{x}^2}{4}-\frac{\text{y}^2}{9}\Big]$
$=\frac{2\text{y}}{3}\Big[\frac{\text{x}^2}{4}+\frac{\text{y}^2}{9}+\frac{2\text{xy}}{6}+\frac{\text{x}^2}{4}+\frac{\text{y}^2}{9}-\frac{2\text{xy}}{6}+\frac{\text{x}^2}{4}-\frac{\text{y}^2}{9}\Big]$
$$By rearranging the variable we get$=\frac{2\text{y}}{3}\Big[\frac{\text{x}^2}{4}+\frac{\text{y}^2}{9}+\frac{\text{x}^2}{4}+\frac{\text{x}^2}{9}\Big]$
$=\frac{2\text{xy}}{3}\Big[\frac{3\text{x}^2}{4}+\frac{\text{y}^2}{9}\Big]$
$=\frac{\text{x}^2\text{y}}{2}+\frac{2\text{y}^3}{27}$
Hence the simplified value of$\Big(\frac{\text{x}}{2}+\frac{\text{y}}{3}\Big)^3-\Big(\frac{\text{x}}{2}-\frac{\text{y}}{3}\Big)^3$ is $\frac{\text{x}^2\text{y}}{2}+\frac{2\text{y}^3}{27}.$
View full question & answer→Question 54 Marks
If $\text{x}-\frac{1}{\text{x}}=5,$ find the value of $\text{x}^3-\frac{1}{\text{x}^3}.$
AnswerGiven,If $\text{x}-\frac{1}{\text{x}}=5$ We know that, $(a - b)^3 = a^3 - b^3 - 3ab(a - b)$ ...(1)
Substitute $\text{x}-\frac{1}{\text{x}}=5$ in eq(1)$\Rightarrow\Big(\text{x}-\frac{1}{\text{x}}\Big)^3=\text{x}^3-\frac{1}{\text{x}^3}-3\Big(\text{x}\times\frac{1}{\text{x}}\Big)\Big(\text{x}-\frac{1}{\text{x}}\Big)$
$\Rightarrow5^3=\text{x}^3-\frac{1}{\text{x}^3}-3\Big(\text{x}-\frac{1}{\text{x}}\Big)$
$\Rightarrow125=\text{x}^3-\frac{1}{\text{x}^3}-(3\times5)$
$\Rightarrow125=\text{x}^3-\frac{1}{\text{x}^3}-15$
$\Rightarrow125+15=\text{x}^3-\frac{1}{\text{x}^3}$
$\Rightarrow\text{x}^3-\frac{1}{\text{x}^3}=140$
Hence, the result is $\text{x}^3-\frac{1}{\text{x}^3}=140.$
View full question & answer→Question 64 Marks
Simplify the following:
$(2 x-5 y)^3-(2 x+5 y)^3$
AnswerGiven $(2 x-5 y)^3-(2 x+5 y)^3$ We shall use the identity $a^3-b^3=(a-b)\left(a^2+b^2+a b\right)$ Here $a=(2 x-5 y), b=(2 x+5 y)$ By applying the identity we get $=(2 x -5 y -2 x +5 y )$
$\quad\left[(2 x-5 y)^2+(2 x+5 y)^2((2 x-5 y) \times(2 x+5 y))\right]$
$=(2 x-5 y-2 x-5 y)[2 x \times 2 x+5 y \times 5 y-2 \times 2 x \times 5 y)$
$\left.+(2 x \times 2 x+5 y \times 5 y+2 \times 2 x \times 5 y)+\left(4 x^2-25 y^2\right)\right]$
$=(-10 y)\left[\left(4 x^2+25 y^2-20 xy\right)\right.$
$\left.+\left(4 x^2+25 y^2+20 xy\right)+4 x^2-25 y^2\right]$
$=(-10 y)\left[4 x^2+25 y^2-20 xy+4 x^2+25 y^2+20 xy+4 x^2-25 y^2\right]$
By rearranging the variable we get, $=(-10 y )\left[4 x ^2+4 x ^2+4 x ^2+25 y ^2\right]$
$=-10 y \times\left[12 x^2+25 y^2\right]$
$=-120 x^2 y-250 y^3$
Hence the value of $(2 x -5 y )^3-(2 x +5 y )^3$ is $-120 x ^2 y -250 y ^3$.
View full question & answer→Question 74 Marks
Find the cube the following binomial expressions:$4-\frac{1}{3\text{x}}$
AnswerGiven,$4-\frac{1}{3\text{x}}$
The above equation is in the form of $(a - b)^3= a^3 - b^3 - 3ab(a - b)$ We know that, $\text{a} = 4, \text{b} =\frac{1}{3\text{x}}$ By using $(a - b)^3$ formula$\Big(4-\frac{1}{3\text{x}}\Big)^3$
$=4^3-\Big(\frac{1}{3\text{x}}\Big)^3-3(4)\Big(\frac{1}{3\text{x}}\Big)\Big(4-\frac{1}{3\text{x}}\Big)$
$=64-\frac{1}{27\text{x}^3}-\frac{12}{3\text{x}}\Big(4-\frac{1}{3\text{x}}\Big)$
$=64-\frac{1}{27\text{x}^3}-\frac{4}{\text{x}}\Big(4-\frac{1}{3\text{x}}\Big)$
$=64-\frac{1}{27\text{x}^3}-\Big(\frac{4}{3\text{x}}\times4\Big)+\Big(\frac{4}{3\text{x}}\times\frac{1}{3\text{x}}\Big)$
$=64-\frac{1}{27\text{x}^3}-\frac{16}{\text{x}}+\Big(\frac{4}{3\text{x}^2}\Big)$
Hence The cube of $\Big(4-\frac{1}{3\text{x}}\Big)^3=64-\frac{1}{27\text{x}^3}-\frac{16}{\text{x}}+\Big(\frac{4}{3\text{x}^2}\Big)$
View full question & answer→Question 84 Marks
If $a+b+c=9$ and $a b+b c+c a=26$, find value of $a^3+b^3+c^3-3 a b c$.
AnswerWe know that
$a^3+b^3+c^3-3 a b c=(a+b+c)\left(a^2+b^2+c^2-a b-b c-c a\right)$
$\Rightarrow a^3+b^2+c^3-3 a b c=(a+b+c)\left[\left(a^2+b^2+c^2\right)-(a b+b c+c a)\right] \ldots$
It follow from the above identity that we require the values of $a+b+c, a^2+b^2+c^2$, and $a b+b c+c a$ to get the value of $a^3+b^3+c^3-3 a b c$
The values of $a+b+c$ and $a b+b c+c a$ are known to $u s$.
So we require the value of $a^2+b^2+c^2$,
Now,
$(a+b+c)^2=a^2+b^2+c^2+2(a b+b c+c a)$
$\Rightarrow(9)^2=a^2+b^2+c^2+2 \times 26[\therefore a+b+c=9 \text { and } a b+b c+c a=26]$
$\Rightarrow 81=a^2+b^2+c^2+52$
$\Rightarrow a^2+b^2+c^2=81-52=29$
Substituting the values of $a^2+b^2+c^2$ in (1), we get,
$a^3+b^3+c^3-3 a b c=9(29-26)(\therefore a+b+c=9 \text { and } a b+b c+c a=26)$
$=9 \times 3$
$=27$
$\therefore a^3+b^3+c^3-3 a b c=27$
View full question & answer→Question 94 Marks
If $\text{x}+\frac{1}{\text{x}}=\sqrt{5}$ find the value of $\text{x}^2+\frac{1}{\text{x}^2}$ and $\text{x}^4+\frac{1}{\text{x}^4}$
AnswerWe have,$\Big(\text{x}+\frac{1}{\text{x}}\Big)^2=\text{x}^2+\frac{1}{\text{x}^2}+2\times\text{x}\times\frac{1}{\text{x}}$
$\Rightarrow\Big(\text{x}+\frac{1}{\text{x}}\Big)^2=\text{x}^2+\frac{1}{\text{x}^2}+2$
$\Rightarrow(\sqrt{5})^2=\text{x}^2+\frac{1}{\text{x}^2}+2$ $\big[\because\text{x}+\frac{1}{\text{x}}=\sqrt{5}\big]$
$\Rightarrow5=\text{x}^2+\frac{1}{\text{x}^2}+2$
$\Rightarrow\ \text{x}^2+\frac{1}{\text{x}^2}=3....(\text{i})$
Now, $\Big(\text{x}^2+\frac{1}{\text{x}^2}\Big)^2=\text{x}^4+\frac{1}{\text{x}^4}+2\times\text{x}^2\times\frac{1}{\text{x}^2}$$\Rightarrow\Big(\text{x}^2+\frac{1}{\text{x}^2}\Big)^2=\text{x}^4+\frac{1}{\text{x}^4}+2$
$\Rightarrow9=\text{x}^4+\frac{1}{\text{x}^4}+2$ $\big[\because\ \text{x}^2+\frac{1}{\text{x}^2}=3\big]$
$\Rightarrow\text{x}^4+\frac{1}{\text{x}^4}=7$
Hence, $\text{x}^2+\frac{1}{\text{x}^2}=3,\ \text{x}^4+\frac{1}{\text{x}^4}=7.$
View full question & answer→Question 104 Marks
If $\text{x}-\frac{1}{\text{x}}=\frac{1}{2},$ then write the value of $4\text{x}^2+\frac{4}{\text{x}^2}.$
AnswerWe have to find the value of $4\text{x}^2+\frac{4}{\text{x}^2}$ Given $\text{x}-\frac{1}{\text{x}}=\frac{1}{2}$ Using identity $(a - b)^2 = a^2 - 2ab + b^2$ Here $\text{a} = \text{x},\ \text{b}=\frac{1}{\text{x}}$$\Big(\text{x}-\frac{1}{\text{x}}\Big)^2=\text{x}^2-2\times\text{x}\times\frac{1}{\text{x}}+\Big(\frac{1}{\text{x}}\Big)^2$
$\Big(\text{x}-\frac{1}{\text{x}}\Big)^2=\text{x}^2-2\times\not\text{x}\times\frac{1}{\not\text{x}}+\frac{1}{\text{x}}\times\frac{1}{\text{x}}$
$\Big(\text{x}-\frac{1}{\text{x}}\Big)^2=\text{x}^2-2+\frac{1}{\text{x}^2}$
By substituting the value of $\text{x}-\frac{1}{\text{x}}=\frac{1}{2}$ we get $\Big(\frac{1}{2}\Big)^2=\text{x}^2+\frac{1}{\text{x}^2}-2$
By transposing -2 to left hand side we get $\frac{1}{4}+2=\text{x}^2+\frac{1}{\text{x}^2}$
By taking least common multiply we get $\frac{1}{4}+\frac{2}{1}=\text{x}^2+\frac{1}{\text{x}^2}$
$\frac{1}{4}+\frac{2}{1}\times\frac{4}{4}=\text{x}^2+\frac{1}{\text{x}^2}$
$\frac{1}{4}+\frac{8}{4}=\text{x}^2+\frac{1}{\text{x}^2}$
$\frac{1+8}{4}=\text{x}^2+\frac{1}{\text{x}^2}$
$\frac{+9}{4}=\text{x}^2+\frac{1}{\text{x}^2}$
By multiplying 4 on both sides we get $4\times\frac{9}{4}=4\text{x}^2+4\times\frac{1}{\text{x}^2}$
$4\times\frac{9}{4}=4\text{x}^2+\frac{4}{\text{x}^2}$
$9 = 4\text{x}^2+\frac{4}{\text{x}^2}$
Hence the value of $4\text{x}^2+\frac{4}{\text{x}^2}$ is 9.
View full question & answer→Question 114 Marks
Prove that $\text{a}^2+\text{b}^2+\text{c}^2-\text{ab}-\text{bc}-\text{ca}$ is always non negetive for all values of a, b and c.
AnswerIn the given problem, we have to prove $\text{a}^2+\text{b}^2+\text{c}^2-\text{ab}-\text{bc}-\text{ca}$ is always non negetive for all a, b , c that is we have to prove that $\text{a}^2+\text{b}^2+\text{c}^2-\text{ab}-\text{bc}-\text{ca}\geq0$ Consider,$\text{a}^2+\text{b}^2+\text{c}^2-\text{ab}-\text{bc}-\text{ca}$
$\text{a}^2+\text{b}^2+\text{c}^2-\text{ab}-\text{bc}-\text{ca}\\=\frac{1}{2}\big(2\text{a}^2+2\text{b}^2+2\text{c}^2-2\text{ab}-2\text{bc}-2\text{ca}\big)$
$=\frac{1}{2}[(\text{a}-\text{b}^2)+(\text{b}-\text{c}^2)+(\text{c}-\text{a}^2)]$
View full question & answer→Question 124 Marks
If $a-b=5$ and $a b=12$, find the value of $a^2+b^2$.
AnswerWe have to find the value $a^2+b^2$
Given $a - b =5, ab =12$
Using identity $(a-b)^2=a^2-2 a b+b^2$
By substituting the value of $a-b=5, a b=12$ we get,
$(5)^2=a^2+b^2-2 \times 12$
$5 \times 5=a^2+b^2-2 \times 12$
By transposing -24 to left hand side we get
$25+24=a^2+b^2$
$49=a^2+b^2$
Hence the value of $a^2+b^2$ is 49 .
View full question & answer→Question 134 Marks
If $\text{x}^2+\frac{1}{\text{x}^2}=51,$ find the value of $\text{x}^3-\frac{1}{\text{x}^3}.$
AnswerGiven, $\text{x}^2+\frac{1}{\text{x}^2}=51$ We know that, $(x - y)^2 = x^2 + y^2 - 2xy$ ...(1) Substitute $\text{x}^2+\frac{1}{\text{x}^2}=51$in eq.(1)$\Big(\text{x}-\frac{1}{\text{x}}\Big)^2=\text{x}^2+\frac{1}{\text{x}^2}-2\times\text{x}\times\frac{1}{\text{x}}$
$\Big(\text{x}-\frac{1}{\text{x}}\Big)^2=\text{x}^2+\frac{1}{\text{x}^2}-2$
$\Big(\text{x}-\frac{1}{\text{x}}\Big)^2=51-2$
$\Big(\text{x}-\frac{1}{\text{x}}\Big)^2=49$
$\Big(\text{x}-\frac{1}{\text{x}}\Big)^2=\sqrt{49}$
$\Big(\text{x}-\frac{1}{\text{x}}\Big)^2=\pm7$
We need to find $\text{x}^3-\frac{1}{\text{x}^3}$ So, $a^3 - b^3 = (a - b)(a^2 + b^2 + ab)$$\text{x}^3-\frac{1}{\text{x}^3}=\Big(\text{x}-\frac{1}{\text{x}}\Big)\Big(\text{x}^2+\frac{1}{\text{x}^2}\Big)+\Big(\text{x}\times\frac{1}{\text{x}}\Big)$
We know that,$\Big(\text{x}-\frac{1}{\text{x}}\Big)=7$ and $\Big(\text{x}^2+\frac{1}{\text{x}^2}\Big)=51$
$\Rightarrow\text{x}^3-\frac{1}{\text{x}^3}=7(51+1)$
$\Rightarrow\text{x}^3-\frac{1}{\text{x}^3}=7(52)$
$\Rightarrow\text{x}^3-\frac{1}{\text{x}^3}=364$
Hence, the value of $\text{x}^3-\frac{1}{\text{x}^3}=364.$
View full question & answer→Question 144 Marks
Simplify the following:
$(x + 3)^3 + (x - 3)^3$
AnswerIn the given problem, we have to simplify equation Given $(x + 3)^3 + (x - 3)^3$ We shall use the identity $a^3 + b^3 = (a + b)(a^2 + b^2 - ab)$ Here a= (x + 3), b = (x - 3) By applying identity we get$=\big(\text{x}+\not3+\text{x}-\not3\big)\big[(\text{x}+3)^2+(\text{x}-3)^2-(\text{x}+3)(\text{x}-3)\big]$
$=2\text{x}\big[\big(\text{x}^2+3^2+2\times\text{x}\times3\big)+\big(\text{x}^2+3^2-2\times\text{x}\times3\big)-(\text{x}^2)-3^2\big]$
$=2\text{x}\big[\big(\text{x}^2+9+6\text{x}\big)+\big(\text{x}^2+9-6\text{x}\big)-\big(\text{x}^2-3^2\big)\big]$
$=2\text{x}\big[\text{x}^2+9+6\text{x}+\text{x}^2+9-6\text{x}-\text{x}^2+9\big]$
$=2\text{x}\big[\text{x}^2+\not\text{x}^2-\not\text{x}^2-6\text{x}+6\text{x}+9+9+9\big]$
$=2\text{x}\big[\text{x}^2+27\big]$
$=2\text{x}^2+54\text{x}$
Hence simplified form of expression $(\text{x}+3)^3+(\text{x}-3)^3$ is $2\text{x}^2+54\text{x}.$
View full question & answer→Question 154 Marks
Simplify the following products:$\Big(\text{m}+\frac{\text{n}}{7}\Big)^3\Big(\text{m}-\frac{\text{n}}{7}\Big)$
AnswerIn the given problem, we have to find product of $\Big(\text{m}+\frac{\text{n}}{7}\Big)^3\Big(\text{m}-\frac{\text{n}}{7}\Big)$ We have been given $\Big(\text{m}+\frac{\text{n}}{7}\Big)^3\Big(\text{m}-\frac{\text{n}}{7}\Big)$ On rearranging we get,$\Big(\text{m}+\frac{\text{n}}{7}\Big)^3\Big(\text{m}-\frac{\text{n}}{7}\Big)=\Big(\text{m}+\frac{\text{n}}{7}\Big)^2\Big(\text{m}+\frac{\text{n}}{7}\Big)\Big(\text{m}-\frac{\text{n}}{7}\Big)$
We shall use the identity $\big(\text{x}-\text{y}\big)\big(\text{x}+\text{y}\big)=\text{x}^2-\text{y}^2$ By substituting $\text{x}=\text{m},\ \text{y}=\frac{\text{n}}{7},$ we get,$\Big(\text{m}+\frac{\text{n}}{7}\Big)^3\Big(\text{m}-\frac{\text{n}}{7}\Big)$
$=\Big(\text{m}+\frac{\text{n}}{7}\Big)^2\bigg(\text{m}^2-\Big(\frac{\text{n}}{7}\Big)^2\bigg)$
$=\Big(\text{m}+\frac{\text{n}}{7}\Big)^2\Big(\text{m}^2-\frac{\text{n}^2}{49}\Big)$
Hence the value of $\Big(\text{m}+\frac{\text{n}}{7}\Big)^3\Big(\text{m}-\frac{\text{n}}{7}\Big)$ is $\Big(\text{m}+\frac{\text{n}}{7}\Big)^2\Big(\text{m}^2-\frac{\text{n}^2}{49}\Big)$
View full question & answer→Question 164 Marks
If $\text{x}-\frac{1}{\text{x}}=7,$ find the value of $\text{x}^3-\frac{1}{\text{x}^3}.$
AnswerGiven,If $\text{x}-\frac{1}{\text{x}}=7$We know that, $(a - b)^3 = a^3 - b^3 + 3ab(a + b)$ ...(1)
Substitute $\text{x}-\frac{1}{\text{x}}=7$ in eq(1)
$\Rightarrow\Big(\text{x}-\frac{1}{\text{x}}\Big)^3=\text{x}^3-\frac{1}{\text{x}^3}-3\Big(\text{x}\times\frac{1}{\text{x}}\Big)\Big(\text{x}-\frac{1}{\text{x}}\Big)$
$\Rightarrow7^3=\text{x}^3-\frac{1}{\text{x}^3}-3\Big(\text{x}\times\frac{1}{\text{x}}\Big)\Big(\text{x}-\frac{1}{\text{x}}\Big)$
$\Rightarrow343=\text{x}^3-\frac{1}{\text{x}^3}-(3\times7)$
$\Rightarrow343=\text{x}^3-\frac{1}{\text{x}^3}-21$
$\Rightarrow125+21=\text{x}^3-\frac{1}{\text{x}^3}$
$\Rightarrow\text{x}^3-\frac{1}{\text{x}^3}=364$
Hence, the result is $\text{x}^3-\frac{1}{\text{x}^3}=364.$
View full question & answer→Question 174 Marks
If $x+y+z=8$ and $x y+y z+z x=20$, find value of $x^3+y^3+z^3-3 x y z$.
AnswerWe know that
$x^3+y^3+z^3-3 x y z=(x+y+z)\left(x^2+y^2+z^2-x y-y z-z x\right)$
$\Rightarrow x^3+y^2+z^3-3 x y z=(x+y+z)\left[\left(x^2+y^2+z^2\right)-(x y+y z+z x)\right]$
It follow from the above identity that we require the values of $x+y+z, x^2+y^2+z^2$, and $x y+y z+z x$ to get the value of $x^3+y^3+z^3-3 x y z$
The values of $x+y+z$ and $x y+y z+z x$ are known to $u s$.
So we require the value of $x^2+y^2+z^2$,
Now,
$(x+y+z)^2=x^2+y^2+z^2+2(x y+y z+z x)$
$\Rightarrow(8)^2=x^2+y^2+z^2+2(20)[\therefore x+y+z=8 \text { and } x y+y z+z x=20]$
$\Rightarrow 64=x^2+y^2+z^2+40$
$\Rightarrow x^2+y^2+z^2=64-40=24$
Substituting the values of $x^2+y^2+z^2, x+y+z$ and $x y+y z+z x$ in equation (1),
we get, $x^3+y^3+z^3-3 x y z=8 \times(24-20)$
$=8 \times 4$
$=32$
$\therefore x^3+y^3+z^3-3 x y z=32$
View full question & answer→Question 184 Marks
If $\text{x}^4+\frac{1}{\text{x}^4}=194,$ find $\text{x}^3+\frac{1}{\text{x}^3},\ \text{x}^2+\frac{1}{\text{x}^2}$ and $\text{x}+\frac{1}{\text{x}}.$
AnswerIn the given problem, we have to find the value of $\text{x}^3+\frac{1}{\text{x}^3},\ \text{x}^2+\frac{1}{\text{x}^2},\ \text{x}+\frac{1}{\text{x}}$ Given $\text{x}^4+\frac{1}{\text{x}^4}=194$ By adding and subtracting $2\times\text{x}^2\times\frac{1}{\text{x}^2}$ in left hand side of $\text{x}^4+\frac{1}{\text{x}^4}=194$ we get,$\text{x}^4+\frac{1}{\text{x}^4}+2\times\text{x}^2\times\frac{1}{\text{x}^2}-2\times\text{x}^2\times\frac{1}{\text{x}^2}=194$
$\text{x}^4+\frac{1}{\text{x}^4}+2\times\text{x}^2\times\frac{1}{\text{x}^2}-2\times\Big(\not\text{x}^2\times\frac{1}{\not\text{x}^2}\Big)=194$
$\Big(\text{x}^2\times\frac{1}{\text{x}^2}\Big)^2-2=194$
$\Big(\text{x}^2\times\frac{1}{\text{x}^2}\Big)^2=194+2$
$\Big(\text{x}^2\times\frac{1}{\text{x}^2}\Big)^2=196$
$\Big(\text{x}^2\times\frac{1}{\text{x}^2}\Big)^2=(14)^2$
$\Big(\text{x}^2\times\frac{1}{\text{x}^2}\Big)=14$
Again by adding and subtracting $2\times\text{x}\times\frac{1}{\text{x}}$ in left hand side of $\Big(\text{x}^2+\frac{1}{\text{x}^2}\Big)=14$ we get,$\text{x}^2+\frac{1}{\text{x}^2}+2\times\text{x}\times\frac{1}{\text{x}}-2\times\text{x}\times\frac{1}{\text{x}}=14$
$\Big(\text{x}+\frac{1}{\text{x}}\Big)^2-2\times\not\text{x}\times\frac{1}{\not\text{x}}=14$
$\Big(\text{x}+\frac{1}{\text{x}}\Big)^2-2=14$
$\Big(\text{x}+\frac{1}{\text{x}}\Big)^2=14+2$
$\Big(\text{x}+\frac{1}{\text{x}}\Big)^2=16$
$\Big(\text{x}+\frac{1}{\text{x}}\Big)^2=4\times4$
$\Big(\text{x}+\frac{1}{\text{x}}\Big)=4$
Now cubing on both sides of $\Big(\text{x}+\frac{1}{\text{x}}\Big)=4$ we get$\Big(\text{x}+\frac{1}{\text{x}}\Big)^3=4^3$
we shall use identity $(a + b)^3 = a^3 + b^3 + 3ab(a + b)$$\text{x}^3+\frac{1}{\text{x}^3}+3\times\text{x}\times\frac{1}{\text{x}}\Big(\text{x}\times\frac{1}{\text{x}}\Big)=4\times4\times4$
$\text{x}^3+\frac{1}{\text{x}^3}+3\times\not\text{x}\times\frac{1}{\not\text{x}}\times4=64$
$\text{x}^3+\frac{1}{\text{x}^3}+12=64$
$\text{x}^3+\frac{1}{\text{x}^3}=64-12$
$\text{x}^3+\frac{1}{\text{x}^3}=52$
Hence the value of $\text{x}^3+\frac{1}{\text{x}^3},\ \text{x}^2+\frac{1}{\text{x}^2} ,\ \text{x}+\frac{1}{\text{x}}$ is 52, 14, 4 respectively.
View full question & answer→Question 194 Marks
If $\text{x}+\frac{1}{\text{x}}=5,$ find the value of $\text{x}^3+\frac{1}{\text{x}^3}.$
AnswerGiven, $\text{x}+\frac{1}{\text{x}}=5$
We know that, $(a + b)^3 = a^3 + b^3 + 3ab(a + b)$...(1)
Substitute $\text{x}+\frac{1}{\text{x}}=5$ in eq(1)
$\Rightarrow\Big(\text{x}+\frac{1}{\text{x}}\Big)^3=\text{x}^3+\frac{1}{\text{x}^3}+3\Big(\text{x}\times\frac{1}{\text{x}}\Big)\Big(\text{x}+\frac{1}{\text{x}}\Big)$
$\Rightarrow5^3=\text{x}^3+\frac{1}{\text{x}^3}+3\Big(\text{x}\times\frac{1}{\text{x}}\Big)\Big(\text{x}+\frac{1}{\text{x}}\Big)$
$\Rightarrow125=\text{x}^3+\frac{1}{\text{x}^3}+3\Big(\text{x}+\frac{1}{\text{x}}\Big)$
$\Rightarrow125=\text{x}^3+\frac{1}{\text{x}^3}+3(5)$
$\Rightarrow125=\text{x}^3+\frac{1}{\text{x}^3}+15$
$\Rightarrow125-15=\text{x}^3+\frac{1}{\text{x}^3}$
$\Rightarrow\text{x}^3+\frac{1}{\text{x}^3}=110$
Hence, the result is $\text{x}^3+\frac{1}{\text{x}^3}=110.$
View full question & answer→Question 204 Marks
Simplify:
$(a + b + c)^2- (a - b + c)^2$
AnswerIn the given problem, we have to simplify the expressions Given $(a + b + c)^2 - (a - b + c)^2$ By using identity $(x + y + z)^2 = x^2 + y^2 + z^2 + 2xy + 2yz + 2zx$
Hence the equation becomes
$\big(\text{a}+\text{b}+\text{c}\big)-\big(\text{a}-\text{b}+\text{c}\big)\\=\Big[\text{a}^2+\text{b}^2+\text{c}^2+2\text{ab}+2\text{bc}+2\text{ca}\Big]\\-\Big[\text{a}^2+(-\text{b})^2+\text{c}^2+2\text{a}(-\text{b})+2(-\text{b})(\text{c})+2\text{ca}\Big]$
$=\text{a}^2+\text{b}^2+\text{c}^2+2\text{ab}+2\text{bc}+2\text{ca}\\-\text{a}^2-\text{b}^2-\text{c}^2+2\text{ab}+2\text{bc}-2\text{ca}$
$=\not\text{a}^2-\not\text{a}^2+\not\text{b}^2-\not\text{b}^2+\not\text{c}^2-\not\text{c}^2\\+2\text{ab}+2\text{ab}+2\text{bc}+2\text{bc}+2\text{ca}-2\text{ca}$
$=4\text{ab}+4\text{bc}$
Talking 4 as common factor we get$=4\big(\text{ab}+\text{bc}\big)$
Hence the simplified value of $\big(\text{a}+\text{b}+\text{c}\big)^2-\big(\text{a}-\text{b}+\text{c}\big)^2$ is $4\big(\text{ab}+\text{bc}\big).$
View full question & answer→Question 214 Marks
If $a+b+c=9$ and $a^2+b^2+c^2=35$, find value of $a^3+b^3+c^3-3 a b c$
AnswerWe know that
$a^3+b^3+c^3-3 a b c=(a+b+c)\left(a^2+b^2+c^2-a b-b c-c a\right)$
$\Rightarrow a^3+b^3+c^3-3 a b c=(a+b+c)\left[\left(a^2+b^2+c^2\right)-(a b+b c+c a)\right] \ldots(1)$
It follow from the above identity that we require the values of $a+b+c, a^2+b^2+c^2$, and $a b+b c+c a$ to get the value of $a^3+b^3+c^3-3 a b c$.
The values of $a+b+c$ and $a^2+b^2+c^2$ are known to us.
So we require the value of $a b+b c+c a$,
Now,
$(a+b+c)^2=a^2+b^2+c^2+2(a b+b c+c a)$
$\Rightarrow(9)^2=35+2(a b+b c+c a)\left[\therefore a+b+c=9 a n d a^2+b^2+c^2=35\right]$
$\Rightarrow 81=35+2(a b+b c+c a)$
$\Rightarrow 2(a b+b c+c a) 81-35=46$
$\Rightarrow a b+b c+c a=\frac{46}{2}=23$
Substituting the values of $a b+b c+c a$ in (1), we get,
$a^3+b^3+c^3-3 a b c=9(35-23)\left(\therefore a+b+c=9 a n d a^2+b^2+c^2=35\right)$
$=9 \times 12$
$=108$
$\therefore a^3+b^3+c^3-3 a b c=108$
View full question & answer→Question 224 Marks
If $2x + 3y = 8$ and $xy = 2$, find the value of $4x^2 + 9y^2.$
AnswerIn the given problem, we have to find $4 x^2+9 y^2$ We have been given $2 x+3 y=8$ and $x y=2$
Let us take $2 x+3 y=8$
On squaring both side we get,
$(2 x+3 y)^2=(8)^2$
We shall use the identity $(x+y)^2=x^2+2 x y+y^2$
$(2 x \times 2 x+3 y \times 3 y+2 \times 2 x \times 3 y)=64$
$4 x^2+9 y^2+12 x y=64$
$4 x^2+9 y^2+12(x y)=64$
By substituting $x y=2$ we get
$4 x^2+9 y^2+12(2)=64$
$4 x^2+9 y^2+24=64$
$4 x^2+9 y^2=64-24$
$4 x^2+9 y^2=40$
Hence the value of $4 x^2+9 y^2$ is 40
View full question & answer→Question 234 Marks
If $\text{a}^2-\frac{1}{\text{a}^2}=102,$ find the value of $\text{a}-\frac{1}{\text{a}}.$
AnswerWe have to find the value of $\text{a}-\frac{1}{\text{a}}$
Given $\text{a}^2-\frac{1}{\text{a}^2}=102$
Using identity $(x - y)^2 = x^2 + y^2 - 2xy$
Here $\text{x}=\text{a},\ \text{y}=\frac{1}{\text{a}}$$\Big(\text{a}-\frac{1}{\text{a}}\Big)^2=\text{a}^2+\Big(\frac{1}{\text{a}}\Big)^2-2\times\text{a}\times\frac{1}{\text{a}}$
$\Big(\text{a}-\frac{1}{\text{a}}\Big)^2=\text{a}^2+\frac{1}{\text{a}^2}-2\times\not\text{a}\times\frac{1}{\not\text{a}}$
By substituting $\text{a}^2-\frac{1}{\text{a}^2}=102$ we get $\Big(\text{a}-\frac{1}{\text{a}}\Big)^2=102-2$
$\Big(\text{a}-\frac{1}{\text{a}}\Big)^2=100$
$\Big(\text{a}-\frac{1}{\text{a}}\Big)\Big(\text{a}-\frac{1}{\text{a}}\Big)=10\times10$
$\Big(\text{a}-\frac{1}{\text{a}}\Big)=10$
Hence the value of $\Big(\text{a}-\frac{1}{\text{a}}\Big)$ is 10.
View full question & answer→Question 244 Marks
Simplify the following expressions:
$(x^2 - x + 1)^2 - (x^2 + x + 1)^2$
AnswerExpanding,
we get $\left[x^2-x+1\right]^2-\left[x^2+x+1\right]^2$
$\left.=\left(x^2\right)^2+(-x)^2+1^2+2\left(x^2\right)(-x)+2(-x)(1)+2 x^2\right)-\left[\left(x^2\right)^2+x^2+1+2 x^2 x+2 x(1)+2 x^2(1)\right]$
${\left[\therefore(x+y+z)^2=x^2+y^2+z^2+2 x y+2 y z+2 x z\right]}$
$=x^4+y^2+1-2 x^3-2 x+2 x^2-x^2-x^4-1-2 x^3-2 x-2 x^2$
$=-4 x^3-4 x=-4 x\left(x^2+1\right)$
Hence simplified equation
$=\left[x^2-x+1\right]^2-\left[x^2+x+1\right]^2$
$=-4 x\left(x^2+1\right)$
View full question & answer→Question 254 Marks
Find the cube of the following binomial expressions:$\frac{1}{\text{x}}+\frac{\text{y}}{3}$
AnswerGiven,$\frac{1}{\text{x}}+\frac{\text{y}}{3}$
The above equation is in the form of $(a + b)^3= a^3 + b^3 + 3ab(a + b)$
We know that,$ \text{a} =\frac{1}{\text{x}},\text{b} =\frac{\text{y}}{3}$
By using $(a + b)^3$ formula$\Big(\frac{1}{\text{x}}+\frac{\text{y}}{3}\Big)^3$
$=\Big(\frac{1}{\text{x}}\Big)^3+\Big(\frac{\text{y}}{3}\Big)^3+3\Big(\frac{1}{\text{x}}\Big)\Big(\frac{\text{y}}{3}\Big)\Big(\frac{1}{\text{x}}+\frac{\text{y}}{3}\Big)$
$=\frac{1}{\text{x}^3}+\frac{\text{y}^3}{27}+3\times\frac{1}{\text{x}}\times\frac{\text{y}}{3}\Big(\frac{1}{\text{x}}+\frac{\text{y}}{3}\Big)$
$=\frac{1}{\text{x}^3}+\frac{\text{y}^3}{27}+\frac{\text{y}}{\text{x}}\Big(\frac{1}{\text{x}}+\frac{\text{y}}{3}\Big)$
$=\frac{1}{\text{x}^3}+\frac{\text{y}^3}{27}+\Big(\frac{\text{y}}{\text{x}}\times\frac{1}{\text{x}}\Big)+\Big(\frac{\text{y}}{\text{x}}\times\text{y}^3\Big)$
$=\Big(\frac{1}{\text{x}^3}+\frac{\text{y}^3}{27}+\frac{\text{y}}{\text{x}^2}+\frac{\text{y}^2}{3\text{x}}\Big)$
Hence$\Big(\frac{1}{\text{x}}+\frac{\text{y}}{3}\Big)^3=\Big(\frac{1}{\text{x}^3}+\frac{\text{y}^3}{27}+\frac{\text{y}}{\text{x}^2}+\frac{\text{y}^2}{3\text{x}}\Big)$
View full question & answer→Question 264 Marks
If $\text{x}+\frac{1}{\text{x}}=3,$ then find the value of $\text{x}^2+\frac{1}{\text{x}^2}.$
AnswerWe have to find the value of $\text{x}^2+\frac{1}{\text{x}^2}$
Given $\text{x}+\frac{1}{\text{x}}=3$
Using identity $(a+b)^2=a^2+2 a b+b^2$
Here $\text{a}=\text{x},\ \text{b}=\frac{1}{\text{x}}$$\Big(\text{x}+\frac{1}{\text{x}}\Big)^2=\text{x}^2+2\times\text{x}\times\frac{1}{\text{x}}+\Big(\frac{1}{\text{x}}\Big)^2$
$\Big(\text{x}+\frac{1}{\text{x}}\Big)^2=\text{x}\times\text{x}+2\times\not{\text{x}}\times\frac{1}{\not{\text{x}}}+\frac{1}{\text{x}}\times\frac{1}{\text{x}}$
$\Big(\text{x}+\frac{1}{\text{x}}\Big)^2=\text{x}^2+2+\frac{1}{\text{x}^2}$
By substituting the value of $\text{x}+\frac{1}{\text{x}}=3$ we get,$(3)^2=\text{x}^2+2+\frac{1}{\text{x}^2}$
$3\times3=\text{x}^2+2+\frac{1}{\text{x}^2}$
By transposing + 2 to left hand side, we get$9-2=\text{x}^2+\frac{1}{\text{x}^2}$
$7=\text{x}^2+\frac{1}{\text{x}^2}$
Hence the value of $\text{x}^2+\frac{1}{\text{x}^2}$ is $7$.
View full question & answer→Question 274 Marks
Simplify:
$(2x + p - c)^2 - (2x - p + c)^2$
AnswerGiven $(2x + p - c)^2 - (2x - p + c)^2$
By using identity $(x + y + z) = x^2 + y^2 + z^2 + 2xy + 2yz + 2zx,$
we get $(2x + p - c)^2 - (2x - p + c)^2$
$=(2\text{x})^2+(\text{p})^2+(-\text{c})^2+2(2\text{x})(\text{p})+2(\text{p})(-\text{c})+2(-\text{c})(2\text{x})\\-\big[(2\text{x})^2+(-\text{p})^2+(\text{c})^2+2(2\text{x})(-\text{p})+2(-\text{p})(\text{c})+2(\text{c})(2\text{x})\big]$
$$$=4\text{x}^2+\text{p}^2+\text{c}^2+4\text{xp}-2\text{cp}-4\text{cx}\\-\big[4\text{x}^2+\text{p}^2+\text{c}^2-4\text{xp}-2\text{cp}+4\text{cx}\big]$
By cancelling the opposite terms,
we get$\big(2\text{x}+\text{p}-\text{c}\big)^2-\big(2\text{x}-\text{p}+\text{c}\big)^2\\=4\text{x}^2+\not\text{p}^2+\not\text{c}^2+4\text{xp}-2\text{cp}-4\text{cx}-4\text{x}^2\\-\not\text{p}^2-\not\text{c}^2+4\text{xp}+2\text{cp}-4\text{cx}$
$=4\text{xp}+4\text{xp}-4\text{cx}-4\text{cx}$
$=8\text{xp}-8\text{cx}$
Talking $8x$ as common a factor we get,$\big(2\text{x}+\text{p}-\text{c}\big)^2-\big(2\text{x}-\text{p}+\text{c}\big)^2=8\text{x}(\text{p}-\text{c})$
Hence the value of $\big(2\text{x}+\text{p}-\text{c}\big)^2-\big(2\text{x}-\text{p}+\text{c}\big)^2$ is $8\text{x}(\text{p}-\text{c}).$
View full question & answer→Question 284 Marks
If $\text{x}^2+\frac{1}{\text{x}^2}=79,$ find the value of $\text{x}+\frac{1}{\text{x}}.$
AnswerIn the given problem, we have to find $\Big(\text{x}+\frac{1}{\text{x}}\Big)$ Given $\text{x}^2+\frac{1}{\text{x}^2}=79$ Adding and subtracting 2 on left hand side,$\text{x}^2+\frac{1}{\text{x}^2}+2-2=79$
$\Big(\text{x}^2+\frac{1}{\text{x}^2}+2\times\text{x}\times\frac{1}{\text{x}}\Big)-2=79$
$\Big(\text{x}+\frac{1}{\text{x}}\Big)^2-2=79$
$\Big(\text{x}+\frac{1}{\text{x}}\Big)^2=79+2$
$\Big(\text{x}+\frac{1}{\text{x}}\Big)^2=81$
$\Big(\text{x}+\frac{1}{\text{x}}\Big)^2=\sqrt{81}$
$\text{x}+\frac{1}{\text{x}}=\sqrt{9\times9}$
$\text{x}+\frac{1}{\text{x}}=\pm9$
Hence the value of $\Big(\text{x}+\frac{1}{\text{x}}\Big)$ is $\pm9$
View full question & answer→Question 294 Marks
Simplify the following products:$\Big(\frac{1}{2}\text{a}-3\text{b}\Big)\Big(3\text{b}+\frac{1}{2}\text{a}\Big)\Big(\frac{1}{4}\text{a}^2+9\text{b}^2\Big)$
AnswerIn the given problem, we have to find product of $\Big(\frac{1}{2}\text{a}-3\text{b}\Big)\Big(3\text{b}+\frac{1}{2}\text{a}\Big)\Big(\frac{1}{4}\text{a}^2+9\text{b}^2\Big)$ We have been given $\Big(\frac{1}{2}\text{a}-3\text{b}\Big)\Big(3\text{b}+\frac{1}{2}\text{a}\Big)\Big(\frac{1}{4}\text{a}^2+9\text{b}^2\Big)$ On rearranging we get, $\Big(\frac{1}{2}\text{a}-3\text{b}\Big)\Big(\frac{1}{2}\text{a}+3\text{b}\Big)\Big(\frac{1}{4}\text{a}^2+9\text{b}^2\Big)$ We shall use the identity $\big(\text{x}-\text{y}\big)\big(\text{x}+\text{y}\big)=\text{x}^2-\text{y}^2$ By substituting $\text{x}=\frac{1}{2}\text{a},\ \text{y}=3\text{b}$ we get,$$$\Big(\frac{1}{2}\text{a}-3\text{b}\Big)\Big(3\text{b}+\frac{1}{2}\text{a}\Big)\Big(\frac{1}{4}\text{a}^2+9\text{b}^2\Big)\\=\Big(\frac{1}{2}\text{a}\Big)^2-(3\text{b})^2\Big(\frac{1}{4}\text{a}^2+9\text{b}^2\Big)$
$=\Big(\frac{1}{4}\text{a}^2-9\text{b}^2\Big)\Big(\frac{1}{4}\text{a}^2+9\text{b}^2\Big)$
We shall use the identity $\big(\text{x}-\text{y}\big)\big(\text{x}+\text{y}\big)=\text{x}^2-\text{y}^2$$\Big(\frac{1}{2}\text{a}-3\text{b}\Big)\Big(3\text{b}+\frac{1}{2}\text{a}\Big)\Big(\frac{1}{4}\text{a}^2+9\text{b}^2\Big)\\=\Big(\frac{1}{4}\text{a}^2\Big)^2-(9\text{b}^2)^2$
$=\frac{1}{16}\text{a}^4-81\text{b}^4$
Hence the value of $\Big(\frac{1}{2}\text{a}-3\text{b}\Big)\Big(3\text{b}+\frac{1}{2}\text{a}\Big)\Big(\frac{1}{4}\text{a}^2+9\text{b}^2\Big)$ is $\frac{1}{16}\text{a}^4-81\text{b}^4.$
View full question & answer→Question 304 Marks
If $a+b=10$ and $a b=16$, find the value of $a^2-a b+b^2$ and $a^2+a b+b^2$.
AnswerWe have,
$a^2-a b+b^2=a^2+b^2-a b$
$=a^2+b^2-a b+2 a b-2 a b[\text { Adding and substracting 2ab] }$
$=\left(a^2+b^2+2 a b\right)-3 a b[\because(a+b) 2=a 2+b 2-2 a b]$
$=(a+b)^2-3 \times 16[\because a+b=10 \text { and } b=16]$
$=(10)^2-3 \times 16$
$=100-48$
$=52$
$\Rightarrow a^2-a b+b^2=52$
We have,
$a^2+a b+b^2=a^2+a b+b^2+a b-a b[\text { Adding and substracting } a b]$
$=\left(a^2+b^2+2 a b\right)-a b$
$=(a+b)^2-a b[\because(a+b) 2=a 2+b 2-2 a b]$
$=(10)^2-16$
$=100-16$
$\Rightarrow a^2+a b+b^2=84$
Hence, $a ^2- ab + b ^2=52$, and $a ^2+ ab + b ^2=84$.
View full question & answer→Question 314 Marks
Find the value of $64 x^3-125 z^3$, if $4 x-5 z=16$ and $x z=12$.
AnswerGiven, $64 x^3-125 z^3$
Here, $4 x-5 z=16$ and $x z=12$
Cubing $4 x-5 z=16$ on both sides
$(4 x-5 z)^3=16^3$
We know that, $(a-b)^3=a^3-b^3-3 a b(a-b)$
$(4 x)^3-(5 z)^3-3(4 x)(5 z)(4 x-5 z)=16^3$
$64 x^3-125 z^3-60(x z)(16)=4096$
$64 x^3-125 z^3-60(12)(16)=4096$
$64 x^3-125 z^3-11520=4096$
$64 x^3-125 z^3=4096+11520$
$64 x^3-125 z^3=15616$
The value of $64 x^3-125 z^3=15616$
View full question & answer→Question 324 Marks
Find the value of $27 x^3+8 y^3$, if
$3 x+2 y=14 \text { and } x y=8$
AnswerIn the given problem, we have to find the value of $27 x^3+8 y^3$
Given $3 x+2 y=14, x y=8$
On cubing both sides we get,
$(3 x+2 y)^3=(14)^3$
We shall use identity $(a+b)^3=a^3+b^3+3 a b(a+b)$
$27 x^3+8 y^3+3(3 x)(2 y)(3 x+2 y)=14 \times 14 \times 14$
$27 x^3+8 y^3+18(x y)(3 x+2 y)=14 \times 14 \times 14$
$27 x^3+8 y^3+18(8)(14)=2744$
$27 x^3+8 y^3+2016=2744$
$27 x^3+8 y^3=2744-2016$
$27 x^3+8 y^3=728$
Hence the value of $27 x^3+8 y^3$ is $728$
View full question & answer→Question 334 Marks
If $2x + 3y = 13$ and $xy = 6,$ find the value of $8x^3 + 27y^3.$
AnswerGiven, $2 x+3 y=13, x y=6$
We know that,
$(2 x+3 y)^3=13^2$
$\Rightarrow 8 x^3+27 y^3+3(2 x)(3 y)(2 x+3 y)=2197$
$\Rightarrow 8 x^3+27 y^3+18 x y(2 x+3 y)=2197$
Substitute $2 x+3 y=13, x y=6$
$\Rightarrow 8 x^3+27 y^3+18(6)(13)=2197$
$\Rightarrow 8 x^3+27 y^3+1404=2197$
$\Rightarrow 8 x^3+27 y^3=2197-1404$
$\Rightarrow 8 x^3+27 y^3=793$
Hence, the value of $8 x^3+27 y^3=793$
View full question & answer→Question 344 Marks
If $3x - 7y = 10$ and $xy = -1$, find the value of $9x^2 + 49y^2.$
AnswerIn the given problem, we have to find $9 x^2+49 y^2$
We have been given $3 x-7 y=10$ and $x y=-1$
Let us take $3 x-7 y=10$
On squaring both side we get,
$(3 x-7 y)^2=(10)^2$
$(3 x \times 3 x+7 y \times 7 y-2 \times 3 x \times 7 y)=100$
We shall use the identity $(x-y)^2=x^2+2 x y+y^2$
$9 x^2+49 y^2-42 x y=100$
$9 x^2+49 y^2-42(x y)=100$
By substituting $x y=-1$ we get
$9 x^2+49 y^2-42(-1)=100$
$9 x^2+49 y^2+42=100$
$9 x^2+49 y^2=100-42$
$9 x^2+49 y^2=58$
Hence the value of $9 x^2+49 y^2$ is $58$
View full question & answer→Question 354 Marks
If $\text{x}-\frac{1}{\text{x}}=3+2\sqrt2,$ find the value of $\text{x}^3-\frac{1}{\text{x}^3}.$
AnswerIn the given problem, we have to find the value of $\text{x}^3-\frac{1}{\text{x}^3}$
Given $\text{x}-\frac{1}{\text{x}}=3+2\sqrt2,$
Cubing on both side $\text{x}-\frac{1}{\text{x}}=3+2\sqrt2,$
we get$\Big(\text{x}-\frac{1}{\text{x}}\Big)^3=\Big(3+2\sqrt2\Big)^3$
We shall use identity $(a + b)^3 = a^3 + b^3 + 3ab(a + b)$$\big(3+2\sqrt2\big)^3$
$=\text{x}^3-\frac{1}{\text{x}^3}-3\times\text{x}\times\frac{1}{\text{x}}\Big(\text{x}-\frac{1}{\text{x}}\Big)$
$3^3+\big(2\sqrt2\big)+3\times3\times2\sqrt2\big(3+2\sqrt2\big)\\=\text{x}^3-\frac{1}{\text{x}^3}-3\times\not\text{x}\times\frac{1}{\not\text{x}}\times\big(3+2\sqrt2\big)$
$27+16\sqrt2+18\sqrt2\big(3+2\sqrt2\big)\\=\text{x}^3-\frac{1}{\text{x}^3}-3\big(3+2\sqrt2\big)$
$27+16\sqrt2+18\sqrt2\times3+18\sqrt2\times2\sqrt2\\=\text{x}^3-\frac{1}{\text{x}^3}-9-6\sqrt2$
$27+16\sqrt2+54\sqrt2+72=\text{x}^3-\frac{1}{\text{x}^3}-9-6\sqrt2$
$27+16\sqrt2+54\sqrt2+72+9+6\sqrt2=\text{x}^3-\frac{1}{\text{x}^3}$
$\big[27+72+9\big]+\big[16\sqrt2+54\sqrt2+6\sqrt2\big]=\text{x}^3-\frac{1}{\text{x}^3}$
$108+76\sqrt2=\text{x}^3-\frac{1}{\text{x}^3}$
Hence the value of $\text{x}^3-\frac{1}{\text{x}^3}$ is $108+76\sqrt2.$
View full question & answer→Question 364 Marks
Find the cube of the following binomial expressions:$2\text{x}+\frac{3}{\text{x}}$
AnswerGiven,$2\text{x}+\frac{3}{\text{x}}$
The above equation is in the form of $(a + b)^3= a^3 + b^3 + 3ab(a + b)$
we know that $\text{a}=2\text{x},\text{b}=\frac{3}{\text{x}}$ By using $(a + b)^3$formula$=8\text{x}^3+\frac{27}{\text{x}^3}+\frac{18\text{x}}{\text{x}}\Big(\frac{2}{\text{x}}+\frac{3}{\text{x}}\Big)$
$=8\text{x}^3+\frac{27}{\text{x}^3}+\frac{18\text{x}}{\text{x}}\Big(2{\text{x}}+\frac{3}{\text{x}}\Big)$
$=8\text{x}^3+\frac{27}{\text{x}^3}+\Big({18}\times2{\text{x}}\Big)+\Big(18\times\frac{3}{\text{x}}\Big)$
$=\Big(8\text{x}^3+\frac{27}{\text{x}^3}+36\times\frac{54}{\text{x}}\Big)$
Hence The cube of $\Big(2\text{x}+\frac{3}{\text{x}}\Big)^3=\Big(8\text{x}^3+\frac{27}{\text{x}^3}+36\times\frac{54}{\text{x}}\Big)$
View full question & answer→Question 374 Marks
Simplify:
$(x^2 + y^2 - z^2)^2 - (x^2 - y^2 + z^2)^2$
AnswerWe have $\left(x^2+y^2-z^2\right)^2-\left(x^2-y^2+z^2\right)^2$
Using formula $(x+y+z)^2=x^2+y^2+z^2+2 x y+2 y z+2 z x$,
we get $\left(x^2+y^2-z^2\right)^2-\left(x^2-y^2+z^2\right)^2$
$=(\text{x}^2)^2+(\text{y}^2)^2+(-\text{z}^2)^2+2(\text{x}^2)(\text{y}^2)+2(\text{y}^2)(-\text{z}^2)+2(-\text{z}^2)(\text{x}^2)\\-\Big[(\text{x}^2)^2+(-\text{y}^2)^2+(\text{z}^2)^2+2(\text{x}^2)(-\text{y}^2)+2(-\text{y}^2)(\text{z}^2)+2(\text{z}^2)(\text{x}^2)\Big]$
$=\text{x}^4+\text{y}^4+\text{z}^4+2\text{x}^2\text{y}^2-2\text{y}^2\text{z}^2-2\text{z}^2\text{x}^2\\-\big[\text{x}^4+\text{y}^4+\text{z}^4-2\text{x}^2\text{y}^2-2\text{y}^2\text{z}^2+2\text{z}^2\text{x}^2\big]$
By canceling the opposite terms,
we get$\big(\text{x}^2 +\text{y}^2 -\text{z}^2\big)^2 -\big(\text{x}^2 -\text{y}^2+\text{z}^2)^2=\not\text{x}^4+\not\text{y}^4+\not\text{z}^4\\+2\text{x}^2\text{y}^2-2\text{y}^2\text{z}^2-2\text{z}^2\text{x}^2\not\text{x}^4-\not\text{y}^4-\not\text{z}^4+2\text{x}^2\text{y}^2+2\text{y}^2\text{z}^2-2\text{z}^2\text{x}^2$
$=4\text{x}^2\text{y}^2-4\text{z}^2\text{x}^2$
Talking $4x^2$ as common factor we get$\big(\text{x}^2 +\text{y}^2 -\text{z}^2\big)^2 -\big(\text{x}^2 -\text{y}^2+\text{z}^2)^2=4\text{x}^2(\text{y}^2-\text{z}^2)$
Hence the value of $\big(\text{x}^2 +\text{y}^2 -\text{z}^2\big)^2 -\big(\text{x}^2 -\text{y}^2+\text{z}^2)^2$ is $4\text{x}^2(\text{y}^2-\text{z}^2).$
View full question & answer→Question 384 Marks
Find the value of $27x^3 + 8y^3$, if:
$3x + 2y = 20$ and $\text{xy}=\frac{14}{9}$
AnswerGiven $3 x+2 y=20$
$xy=\frac{14}{9}$
On cubing both sides we get, $(3 x+2 y)^3=(20)^3$ We shall use identity $(a+b)^3$
$=a^3+b^3+3 a b(a+b) 27 x^3+8 y^3+3(3 x)(2 y)(3 x+2 y)$
$=20 \times 20 \times 2027 x^3+8 y^3+18(x y)(3 x+2 y)$
$=800027 x^3+8 y^3+18\left(\frac{14}{9}\right)(20)=8000$
$27 x^3+8 y^3+560=800027 x^3+8 y^3$
$=8000-56027 x^3+8 y^3$
$=7440$
Hence the value of $27 x^3+8 y^3$ is 7440 .
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