Question 13 Marks
Evaluate the following:
$(98)^3$
AnswerWe know that $(a-b)^3=a^3-b^3-3 a b(a-b)$
$\Rightarrow(98)^3$ can be written as $(100-2)^3$
Here, $\mathrm{a}=100$ and $\mathrm{b}=2$
$(98)^3=(100-2)^3$
$=(100)^3-(2)^3-3(100)(2)(100-2)$
$=1000000-8-(600 \times 102)$
$=1000000-8-58800$
$=941192$
The value of $(98)^3=941192$
View full question & answer→Question 23 Marks
Evaluate the following using identities:
$991 \times 1009$
AnswerWe have,
$991 \times 1009$
$=(1000-9)(1000+9)$
$=(1000)^2-(9)\left[(a+b)(a-b)=a^2-b^2\right]$
$=1000000-81[\text { Where } a=1000 \text { and } b=9]$
$=999919$
Therefore, $991 \times 1009=999919$
View full question & answer→Question 33 Marks
Find the following products:
$(4 x-3 y+2 z)\left(16 x^2+9 y^2+4 z^2+12 x y+6 y z-8 z x\right)$
AnswerWe have,
$(4 x-3 y+2 z)\left(16 x^2+9 y^2+4 z^2+12 x y+6 y z-8 z x\right)$
$=(4 x+(-3 y)+2 z)\left[(4 x)^2+(-3 y)^2+(2 z)^2-(4 x)(-3 y)-(-3 y)(2 z)-(2 z)(4 x)\right]$
$=(4 x)^3+(-3 y)^3+(2 z)^3-3 x(4 x) \times(-3 y)(2 z)\left[\because a^3+b^3+c^3=3 a b c=(a+b+c)\left(a^2+b^2+c^2-\right.\right.$
$a b-b c-c a)]$
$=64 x^3-27 y^3+8 z^3+72 x y z$
$\therefore(4 x-3 y+2 z)\left(16 x^2+9 y^2+4 z^2+12 x y+6 y z-8 z x\right)=64 x^3-27 y^3+8 z^3+72 x y z$
View full question & answer→Question 43 Marks
Evaluate: $\Big(\frac{1}{2}\Big)^3+\Big(\frac{1}{3}\Big)^3=\Big(\frac{5}{6}\Big)^3$
AnswerLet $\text{a}=\frac{1}{2},\ \text{b}=\frac{1}{3}$ and $\text{c}=\frac{-5}{6}$ Then,
$\text{a} + \text{b} + \text{c}=\frac{1}{2}+\frac{1}{3}-\frac{5}{6}$
$=\frac{3+2}{6}-\frac{5}{6}$
$\Rightarrow\text{a}+\text{b}+\text{c}=\frac{5}{6}-\frac{5}{6}=0$
$\therefore\text{a}^3+\text{b}^3+\text{c}^3=3\text{abc}$
$\Rightarrow\Big(\frac{1}{2}\Big)^3+\Big(\frac{1}{3}\Big)^3+\Big(\frac{-5}{6}\Big)^3=3\times\Big(\frac{1}{2}\Big)\times\Big(\frac{1}{3}\Big)\times\Big(\frac{-5}{6}\Big)$
$=\frac{-5}{12}$
$\therefore\Big(\frac{1}{2}\Big)^3+\Big(\frac{1}{3}\Big)^3-\Big(\frac{5}{6}\Big)^3=\frac{-5}{12}$
View full question & answer→Question 53 Marks
If $9 x^2+25 y^2=181$ and $x y=-6$, find the value of $3 x+5 y$.
AnswerWe have,
$(3 x+5 y)^2=(3 x)^2+(5 y)^2+2 \times 3 x \times 5 y$
$\Rightarrow(3 x+5 y)^2=9 x^2+25 y^2+30 x y$
$=181+30(-6)\left[\text { Since, } 9 x^2+25 y^2=181 \text { and } x y=-6\right]$
$\Rightarrow(3 x+5 y)^2=1$
$\Rightarrow(3 x+5 y)^2=( \pm 1) 2$
$\Rightarrow 3 x+5 y= \pm 1$
View full question & answer→Question 63 Marks
If $\text{x}^2+\frac{1}{\text{x}^2}=66,$ find the value of $\text{x}-\frac{1}{\text{x}}.$
AnswerWe have, $\Big(\text{x}-\frac{1}{\text{x}}\Big)^2=\text{x}^2+\Big(\frac{1}{\text{x}}\Big)^2-2\times\text{x}\times\frac{1}{\text{x}}$
$\Rightarrow\Big(\text{x}-\frac{1}{\text{x}}\Big)^2=\text{x}^2+\frac{1}{\text{x}^2}-2$
$\Rightarrow\Big(\text{x}-\frac{1}{\text{x}}\Big)^2=66-2$
$\big[\therefore\ \text{x}^2+\frac{1}{\text{x}^2}=66\big]$
$\Rightarrow\Big(\text{x}-\frac{1}{\text{x}}\Big)^2=64$
$\Rightarrow\Big(\text{x}-\frac{1}{\text{x}}\Big)^2=(\pm8)^2$
$\Rightarrow\Big(\text{x}-\frac{1}{\text{x}}\Big)=\pm8$
View full question & answer→Question 73 Marks
If $3 x-2 y=11$ and $x y=12$, find the value of $27 x^3-8 y^3$
AnswerGiven, $3 x-2 y=11, x y=12$
We know that $(a-b)^3=a^3-b^3-3 a b(a+b)$
$(3 x-2 y)^3=11^3$
$\Rightarrow 27 \mathrm{x}^3-8 \mathrm{y}^3-(18 \times 12 \times 11)=1331$
$\Rightarrow 27 \mathrm{x}^3-8 \mathrm{y}^3-2376=1331$
$\Rightarrow 27 x^3-8 y^3=1331+2376$
$\Rightarrow 27 \mathrm{x}^3-8 \mathrm{y}^3=3707$
Hence, the value of $27 x^3-8 y^3=3707$
View full question & answer→Question 83 Marks
If $a-b=4$ and $a b=21$, find the value of $a^3-b^3$
AnswerIn the given problem, we have to find the value of $a^3-b^3$
Given $\mathrm{a}-\mathrm{b}=4, \mathrm{ab}=21$
We shall use the identity $(a-b)^3=a^3-b^3-3 a b(a-b)$
Here putting $\mathrm{a}-\mathrm{b}=4, \mathrm{ab}=21$,
$\Rightarrow(4)^3=a^3-b^3+3(21)(4)$
$\Rightarrow 64=a^3-b^3-252$
$\Rightarrow 64-252=a^3-b^3$
$\Rightarrow 316=a^3-b^3$
Hence, the value of $a^3-b^3$ is $316$ .
View full question & answer→Question 93 Marks
If $x = -2$ and $y = 1$, by using an identity find the value of the following: $\Big(5\text{y}+\frac{15}{\text{y}}\Big)\Big(25\text{y}^2-75+\frac{225}{\text{y}^2}\Big)$
AnswerWe have,$\Big(5\text{y}+\frac{15}{\text{y}}\Big)\Big(25\text{y}^2-75+\frac{225}{\text{y}^2}\Big)$
$\Big(5\text{y}+\frac{15}{\text{y}}\Big)\bigg[\Big(5\text{y})^2-5\text{y}\times\frac{15}{\text{y}}\Big(\frac{15}{\text{y}}\Big)^2\bigg]$
$=(5\text{y})^3+\Big(\frac{15}{\text{y}}\Big)^3$ $\big[\therefore\text{a}^3+\text{b}^3=(\text{a}+\text{b})(\text{a}^2-\text{ab}+\text{b}^2)\big]$
$=125\text{y}^3+\frac{3375}{\text{y}^3}$
$=125(1)^3+\frac{3375}{(1)^3}$ $\big[\therefore\text{y}=1\big]$
$=125+3375$
$=3500$
$\therefore\Big(5\text{y}+\frac{15}{\text{y}}\Big)\Big(25\text{y}^2-75+\frac{225}{\text{y}^2}\Big)=3500$
View full question & answer→Question 103 Marks
If $x=-2$ and $y=1$, by using an identity find the value of the following:
$\left(4 y^2-9 x^2\right)\left(16 y^4+36 x^2 y^2+81 x^4\right)$
AnswerWe have,
$\left(4 y^2-9 x^2\right)\left(16 y^4+36 x^2 y^2+81 x^4\right)$
$=\left(4 y^2-9 x^2\right)\left[\left(4 y^2\right)^2+4 y^2 \times 8 x^2+\left(9 x^2\right)^2\right]$
$=\left(4 y^2\right)^3-\left(9 x^2\right)^3\left[\because a^3-b^3=(a-b)\left(a^2+a b+b^2\right)\right]$
$=64 y^6-729 x^6$
$=64(1)^6-729(-2)^6[\because x=2 \text { and } y=1]$
$=64-729 \times 64$
$=64-46656$
$=-465992$
$\therefore\left(4 y^2-9 x^2\right)\left(16 y^4+36 x^2 y^2+81 x^4\right)=-465992$
View full question & answer→Question 113 Marks
If $a+b+c=0$ and $a^2+b^2+c^2=16$, find the value of $a b+b c+c a$.
AnswerWe know that,
${\left[\therefore(a+b+c)^2=a^2+b^2+c^2+2 a b+2 b c+2 c a\right]}$
$(0)^2=16+2(a b+b c+c a)$
$2(a b+b c+c a)=-16$
$a b+b c+c a=-8$
Hence, value of required express $a b+b c+c a=-8$
View full question & answer→Question 123 Marks
If $x = 3$ and $y = -1$, find the values of the following using in identity: $\Big(\frac{5}{\text{x}}+5\text{x}\Big)\Big(\frac{25}{\text{x}^2}-25+25\text{x}^2\Big)$
AnswerWe have, $\Big(\frac{5}{\text{x}}+5\text{x}\Big)\Big(\frac{25}{\text{x}^2}-25+25\text{x}^2\Big)$
$=\Big(\frac{5}{\text{x}}+5\text{x}\Big)\bigg[\Big(\frac{5}{\text{x}}\Big)^2-\frac{5}{\text{x}}\times5\text{x}+(5\text{x})^2\bigg]$
$=\Big(\frac{5}{\text{x}}\Big)^3+(5\text{x})^3$
$\big[\because\text{a}^3+\text{b}^3=(\text{a}+\text{b})(\text{a}^2-\text{ab}+\text{b}^2)\big]$
$=\frac{125}{\text{x}^3}+125\text{x}^3$
$=\frac{125}{(3)^3}+125(3)^3$
$\big[\because\text{x}=3\big]$
$=\frac{125}{27}+125\times27$
$=\frac{125}{27}+3375$
$=\frac{125+3375\times27}{27}=\frac{125+91125}{27}$
$=\frac{91250}{27}$
View full question & answer→Question 133 Marks
If $a+b=10$ and $a b=21$, find the value of $a^3+b^3$
AnswerGiven,
$a+b=10, a b=21$
We know that, $(a+b)^3=a^3+b^3+3 a b(a+b) \ldots 1$
Substitute $a+b=10, a b=21$ in eq. $1$
$\Rightarrow(10)^3=a^3+b^3+3(21)(10)$
$\Rightarrow 1000=a^3+b^3+630$
$\Rightarrow 1000-630=a^3+b^3$
$\Rightarrow 370=a^3+b^3$
Hence, the value of $a^3+b^3=370$
View full question & answer→Question 143 Marks
Evaluate the following:
$(9.9)^3$
AnswerWe know that $(a-b)^3=a^3-b^3-3 a b(a-b)$
$\Rightarrow(9.9)^3$ can be written as $(10-0.1)^3$
Here, $\mathrm{a}=10$ and $\mathrm{b}=0.1$
$(9.9)^3=(10-0.1)^3$
$=(10)^3-(0.1)^3-3(10)(0.1)(10-0.1)$
$=1000-0.001-(3 \times 9.9)$
$=1000-0.001-29.7$
$=1000-29.701$
$=970.299$
The value of $(9.9)^3=970.299$
View full question & answer→Question 153 Marks
If $x = 3$ and $y = -1$, find the values of the following using in identity:
$\Big(\frac{\text{x}}{7}+\frac{\text{y}}{3}\Big)\Big(\frac{\text{x}^2}{49}+\frac{\text{y}^2}{9}-\frac{\text{xy}}{21}\Big)$
AnswerWe have, $\Big(\frac{\text{x}}{7}+\frac{\text{y}}{3}\Big)\Big(\frac{\text{x}^2}{49}+\frac{\text{y}^2}{9}-\frac{\text{xy}}{21}\Big)$
$=\Big(\frac{\text{x}}{7}+\frac{\text{y}}{3}\Big)\bigg[\Big(\frac{\text{x}}{7}\Big)^2+\Big(\frac{\text{y}}{3}\Big)-\frac{\text{x}}{7}\times\frac{\text{y}}{3}\bigg]$
$=\Big(\frac{\text{x}}{7}\Big)^3+\Big(\frac{\text{y}}{3}\Big)^3$
$\big[\because\text{a}^3+\text{b}^3=(\text{a}+\text{b})(\text{a}^2-\text{ab}+\text{b}^2)\big]$
$=\frac{\text{x}^3}{343}+\frac{\text{y}^3}{27}$
$=\frac{(3)^3}{343}+\frac{(-1)^3}{27}$
$\big[\because\text{x}=3\ \text{and}\ \text{y}=-1\big]$
$=\frac{27}{343}+\frac{-1}{27}$
$=\frac{729-343}{9261}=\frac{386}{9261}$
$\therefore\Big(\frac{\text{x}}{7}+\frac{\text{y}}{3}\Big)\Big(\frac{\text{x}^2}{49}+\frac{\text{y}^2}{9}-\frac{\text{xy}}{21}\Big)=\frac{386}{9261}$
View full question & answer→Question 163 Marks
Evaluate the following:
$(103)^3$
Answer$\text { We know that }(a+b)^3=a^3+b^3+3 a b(a+b)$
$\Rightarrow(103)^3 \text { can be written as }(100+3)^3$
$\text { Here, } a=100 \text { and } b=3$
$(103)^3=(100+3)^3$
$=(100)^3+(3)^3+3(100)(3)(100+3)$
$=1000000+27+(900 \times 103)$
$=1000000+27+92700$
$=1092727$
The value of $(103)^3=1092727$
View full question & answer→Question 173 Marks
Evaluate the following:
$104^3+96^3$
AnswerGiven,
$104^3+96^3$
the above equation can be written as $(100+4)^3+(100-4)^3$
we know that, $(a+b)^3+(a-b)^3=2\left[a^3+3 a b^2\right]$
here, $a=100, b=4$
$(100+4)^3-(100-4)^3=2\left[100^3+3(4)^2(100)\right]$
$=2[1000000+4800]$
$=2[1004800]$
$=200960$
The value of $104^3+96^3=2009600$
View full question & answer→Question 183 Marks
If $a-b=6$ and $a b=20$, find the value of $a^3-b^3$.
AnswerWe have,
$a^3-b^3=(a-b)\left(a^2+a b+b^2\right)$
$=(a-b)\left(a^2+a b+b^2-2 a b+2 a b\right) \text { [Adding and substracting } 2 a b \text { in the second break] }$
$=(a-b)\left[\left(a^2+b^2-2 a b\right)+3 a b\right]$
$=(a-b)\left[(a-b)^2+3 a b\right]\left[\because(a-b)^2=a^2+b^2-2 a b\right]$
$=6 \times\left[(6)^2+3 \times 20\right][\because a-b=6 \text { and } a b=20]$
$=6 \times[36+60]$
$=6 \times 96$
$=576$
$\therefore a^3-b^3=576$
View full question & answer→Question 193 Marks
Simplify the following products:
$(\text{x}^3 -3\text{x}^2-\text{x})(\text{x}^2-3\text{x} + 1)$
Answer In the given problem, we have to find product of
$(\text{x}^3 -3\text{x}^2-\text{x})(\text{x}^2-3\text{x} + 1)$
taking x as common factor $\text{x}(\text{x}^2 -3\text{x}-1)(\text{x}^2-3\text{x} + 1)$
$(\text{x}^3 -3\text{x}^2-\text{x})(\text{x}^2-3\text{x} + 1)\\=\Big[\text{x}\big(\text{x}^2-3\text{x}-1\big)\big(\text{x}^2-3\text{x}+1\big)\Big]$
$=\text{x}\Big[\big\{\big(\text{x}^2-3\text{x}\big)-1\big\}\big\{\big(\text{x}^2-3\text{x}\big)+1\big\}\Big]$
We shall use the identity $(\text{x}-\text{y})(\text{x}+\text{y})=\text{x}^2-\text{y}^2$
$\big(\text{x}^3-3\text{x}^2-\text{x}\big)\big(\text{x}^2-3\text{x}+1\big)\\=\text{x}\Big[\big(\text{x}^2-3\text{x}\big)^2-1^2\Big]$
$=\text{x}\big(\text{x}^4-6\text{x}^3+9\text{x}^2-1\big)$
$=\text{x}^5-6\text{x}^4+9\text{x}^3-\text{x}$
Hence the value of $(\text{x}^3 -3\text{x}^2-\text{x})(\text{x}^2-3\text{x} + 1)$ is $\text{x}^5-6\text{x}^4+9\text{x}^3-\text{x}.$
View full question & answer→Question 203 Marks
Find the following products:
$(3 x+2 y+2 z)\left(9 x^2+4 y^2+4 z^2-6 x y-4 y z-6 z x\right)$
AnswerWe have,
$(3 x+2 y+2 z)\left(9 x^2+4 y^2+4 z^2-6 x y-4 y z-6 z x\right)$
$=(3 x+2 y+2 z)\left((3 x)^2+(2 y)^2+(2 z)^2-3 x \times 2 y \times 2 z-2 z \times 3 x\right)$
$=(3 x)^3+(2 y)^3+(2 z)^3-3 \times 3 x \times 2 y \times 2 z\left[\because a^3+b^3+c^3=3 a b c=(a+b+c)\left(a^2+b^2+c^2-a b-\right.\right.$
$b c-c a)]$
$=27 x^3+8 y^3+8 z^3-36 x y z$
$\therefore(3 x+2 y+2 z)\left(9 x^2+4 y^2+4 z^2-6 x y-4 y z-6 z x\right)=27 x^3+8 y^3+8 z^3-36 x y z$
View full question & answer→Question 213 Marks
Evaluate the following:
$(598)^3$
AnswerWe know that $(a-b)^3=a^3-b^3-3 a b(a-b)$
$\Rightarrow(598)^3 \text { can be written as }(600-2)^3$
$\text { Here, } \mathrm{a}=600 \text { and } \mathrm{b}=2$
$(598)^3=(600-2)^3$
$=(600)^3-(2)^3-3(600)(2)(600-2)$
$=216000000-8-(3600 \times 598)$
$=216000000-8-2152800$
$=216000000-2152808$
$=213847192$
The value of $(598)^3=213847192$
View full question & answer→Question 223 Marks
If $x = 3$, find the values of the following using in identity: $\Big(\frac{3}{\text{x}}-\frac{\text{x}}{3}\Big)\Big(\frac{\text{x}^2}{9}+\frac{9}{\text{x}^2}+1\Big)$
AnswerWe have, $\Big(\frac{3}{\text{x}}-\frac{\text{x}}{3}\Big)\Big(\frac{\text{x}^2}{9}+\frac{9}{\text{x}^2}+1\Big)$
$=\Big(\frac{3}{\text{x}}-\frac{\text{x}}{3}\Big)\bigg[\Big(\frac{\text{x}}{3}\Big)+\Big(\frac{3}{\text{x}}\Big)^2+\frac{3}{\text{x}}\times\frac{\text{x}}{3}\bigg]$
$=\Big(\frac{3}{\text{x}}\Big)^3-\Big(\frac{\text{x}}{3}\Big)^3$
$\big[\because\text{x}^3-\text{y}^3=(\text{x}-\text{y})(\text{x}^2+\text{y}^2+\text{xy})\big]$
$=\frac{27}{\text{x}^3}-\frac{\text{x}^3}{27}$
$=\frac{27}{(3)^3}-\frac{(3)^3}{27}$
$\big[\because\text{x}=3\big]$
$=\frac{27}{27}-\frac{27}{27}$
$=1-1$
$=0$
$\therefore\Big(\frac{3}{\text{x}}-\frac{\text{x}}{3}\Big)\Big(\frac{\text{x}^2}{9}+\frac{9}{\text{x}^2}+1\Big)=0$
View full question & answer→Question 233 Marks
Evaluate the following:
$93^3-107^3$
AnswerGiven,
$93^3-107^3$
the above equation can be written as $(100-7)^3-(100+7)^3$ we know that, $(a-b)^3-(a+b)^3=-2\left[b^3+3 b a^2\right]$
here, $a=93, b=107$
$(100-7)^3-(100+7)^3=-2\left[7^3+3(100)^2(7)\right]$
$=-2[343+210000]$
$=-2[210343]$
$=-420686$
The value of $93^3-107^3=-420686$
View full question & answer→Question 243 Marks
Find the following products:
$(3 x-4 y+5 z)\left(9 x^2+16 y^2+25 z^2+12 x y-15 z x+20 y z\right)$
Answer$(3 x-4 y+5 z)\left(9 x^2+16 y^2+25 z^2+12 x y-15 z x+20 y z\right)$
$=(3 x+(-4 y)+5 z)((3 x) 2+(-4 y) 2+(5 z) 2-(3 x)(-4 y)-(5 z)(3 x))$
$=(3 x)^3+(-4 y)^3+(5 z)^3-3(3 x)(-4 y)(5 z)$
${\left[\because a^3+b^3+c^3-3 a b c=(a+b+c)\left(a^2+b^2+c^2-a b-b c-c a\right)\right]}$
$=27 x^3-64 y^3+125 z^3+180 x y z$
$\therefore(3 x-4 y+5 z)\left(9 x^2+16 y^2+25 z^2+12 x y-15 z x+20 y z\right)=27 x^3-64 y^3+125 z^3+180 x y z$
View full question & answer→Question 253 Marks
If $x = 3$ and $y = -1$, find the values of the following using in identity:
$\left(9 y^2-4 x^2\right)\left(81 y^4+36 x^2 y^2+16 x^4\right)$
AnswerWe have,
$\left(9 y^2-4 x^2\right)\left(81 y^4+36 x^2 y^2+16 x^4\right)$
$=\left(9 y^2-4 x^2\right)\left[\left(9 y^2\right)^2+9 y^2 \times 4 x^2+\left(4 x^2\right)^2\right]$
$=\left(9 y^2\right)^3-\left(4 x^2\right)^3\left[\because a^3-b^3=(a-b)\left(a^2+a b+b^2\right)\right]$
$=729 y^6-64 x^6$
$=729 \times(-1)^6-64(3)^6[\because x=3 \text { and } y=-1]$
$=729-64 \times 729$
$=729-46656$
$=-45927$
$\left(9 y^2-4 x^2\right)\left(81 y^4+36 x^2 y^2+16 x^4\right)=-45927$
View full question & answer→Question 263 Marks
Evaluate the following:
$111^3-89^3$
AnswerGiven,
$111^3-89^3$
the above equation can be written as $(100+11)^3-(100-11)^3$
we know that, $(a+b)^3-(a-b)^3=2\left[b^3+3 a b^2\right]$
here, $\mathrm{a}=100 \mathrm{~b}=11$
$(100+11)^3-(100-11)^3=2\left[11^3+3(100)^2(11)\right]$
$=2[1331+330000]$
$=2[331331]$
$=662662$
The value of $111^3-89^3=662662$
View full question & answer→Question 273 Marks
If $a^2+b^2+c^2=16$ and $a b+b c+c a=10$, find the value of $a+b+c$
AnswerWe know that,
$(a+b+c)^2=a^2+b^2+c^2+2(a b+b c+c a)$
$(x+y+z)^2=16+2(10)$
$(x+y+z)^2=36$
$(x+y+z)=\sqrt{36}$
$(x+y+z)= \pm 6$
Hence, value of required expression $(\mathrm{a}+\mathrm{b}+\mathrm{c})= \pm 6$
View full question & answer→Question 283 Marks
Simplify the following products: $(\text{x}^2 +\text{x}- 2)(\text{x}^2-\text{x} + 2)$
AnswerIn the given problem,
we have to find product of $(\text{x}^2 +\text{x}- 2)(\text{x}^2-\text{x} + 2)$
On rearranging we get $(\text{x}^2 +\text{x}- 2)(\text{x}^2-\text{x} + 2)$
$=\big[\text{x}^2+(\text{x}-2)\big]\big[\text{x}^2-(\text{x}-2)\big]$'
We shall use the identity $(\text{x}-\text{y})(\text{x}+\text{y})=\text{x}^2-\text{y}^2$
$(\text{x}^2 +\text{x}- 2)(\text{x}^2-\text{x} + 2)$
$=\Big[+(\text{x}^2)^2-(\text{x}-2)^2\Big]$
$=\text{x}^4-\big(\text{x}^2-2\times2\times\text{x}+2^2\big)$
$=\text{x}^4-\big(\text{x}^2-4\text{x}+4\big)$
$=\text{x}^4-\text{x}^2+4\text{x}-4$
Hence the value of $(\text{x}^2 +\text{x}- 2)(\text{x}^2-\text{x} + 2)$ is $\text{x}^4-\text{x}^2+4\text{x}-4.$
View full question & answer→Question 293 Marks
Find the value of $4 x^2+y^2+25 z^2+4 x y-10 y z-20 z x$ when $x=4, y=3$ and $z=2$
Answer$4x^2 + y^2 + 25z^2 + 4xy - 10yz - 20zx$
$= (2x)^2 + y^2 + (-5z)^2 + 2(2x)(y) + 2(y)(-5z) + 2(-5z)(2x)$
$= (2x + y - 5z)^2$
$= (2(4) + 3 - 5(2))^2$
$= (8 + 3 - 10)^2$
$= (1)^{2$
$} = 1$
Hence value of the equation is equals to $1$
View full question & answer→Question 303 Marks
Evaluate the following:
$(10.4)^3$
AnswerWe know that $(a+b)^3=a^3+b^3+3 a b(a+b)$
$\Rightarrow(10.4)^3$ can be written as $(10+0.4)^3$
Here, $\mathrm{a}=10$ and $\mathrm{b}=0.4$
$(10.4)^3=(10+0.4)^3$
$=(10)^3+(0.4)^3+3(10)(0.4)(10+0.4)$
$=1000+0.064+(12 \times 10.4)$
$=1000+0.064+124.8$
$=1000+124.864$
$=1124.864$
The value of $(10.4)^3=1124.864$
View full question & answer→Question 313 Marks
If $a+b+c=9$ and $a b+b c+c a=23$, find the value of $a^2+b^2+c^2$.
AnswerWe know that, $(a + b + c)^2$
$= a^2 + b^2 + c^2 + 2(ab + bc + ca) 9^2$
$= a^2 + b^2 + c^2 + 2(23) 81$
$= a^2 + b^2 + c^2 + 46 a^2 + b^2 + c^2$
$= 81 - 46 a^2 + b^2 + c^2$
$= 35$
Hence, value of required expression
$a^2 + b^2 + c^2 = 35$
View full question & answer→Question 323 Marks
Evaluate the following: $46^3 + 34^3$
AnswerGiven, $46^3+34^3$
The above equation can be written as
$(40+6)^3+(40-6)^3$
We know that, $(a+b)^3+(a-b)^3=2\left[a^3+3 a b^2\right]$
here, $a=40, b=4(40+6)^3+(40-6)^3=2\left[40^3+3(6)^2(40)\right]$
$=2[64000+4320]=2[68320]=1366340$
The value of $46^3+34^3=1366340$
View full question & answer→Question 333 Marks
If $a+b=7$ and $a b=12$, find the value of $a^2+b^2$
AnswerWe have to find the value of $a^2+b^2$
Given $\mathrm{a}+\mathrm{b}=7, \mathrm{ab}=12$
Using identity $(a+b)^2=a^2+2 a b+b^2$
By substituting the value of $a+b=7, a b=12$ we get
$(a+b)^2=a^2+b^2+2 \times a b$
$(7)^2=a^2+b^2+2 \times 12$
$49=a^2+b^2+24$
By transposing $+24$ to left hand side we get,
$49-24=a^2+b^2$
$25=a^2+b^2$
Hence the value of $a^2+b^2$ is $25$ .
View full question & answer→Question 343 Marks
If $a + b = 8$ and $ab = 6$, find the value of $a^3 + b^3.$
AnswerWe have, $a^3+b^3=(a+b)\left(a^2-a b+b^2\right)=(a+b)\left(a^2+b^2-a b\right)=(a+b)\left(a^2+b^2-a b+2 a b-2 a b\right)$
[Adding and substracting $2 a b$ in the second break $]
=(a+b)\left[\left(a^2+b^2+2 a b\right)-3 a b\right]=(a+b)\left[(a+b)^2-3 a b\right]\left[\because(a+b)^2=a^2+\right.$
$\left.b^2+2 a b\right]=8 \times\left[(8)^2-3 \times 6\right][\because a+b=8$ and $a b=6]=8 \times[64-18]=8 \times 46=368 \therefore a^3+b^3=368$
View full question & answer→Question 353 Marks
Find the following products: $(2a - 3b - 2c)(4a^2 + 9b^2 + 4c^2 + 6ab - 6bc + 4ca)$
AnswerWe have, $(2 a-3 b-2 c)\left(4 a^2+9 b^2+4 c^2+6 a b-6 b c+4 c a\right) $
$=(2 a+(-3 b)+(-2 c))+\left((2 a)^2+(-3 b)^2+(-2 c)^2-(2 a)(-3 b)(-2 c)-(-2 c)(2 a)\right)$
$=(2 a)^3+(-3 b)^3+(-2 c)^3-3(2 a)(-3 b)(-2 c)\left[\because a^3+b^3+c^3-3 a b c=(a+b+c)\left(a^2+b^2+c^2-a b-b c-c a\right)\right]$
$=8 a^3-27 b^3-8 c^3-36 a b c$
$\therefore(2 a-3 b-2 c)\left(4 a^2+9 b^2+4 c^2+6 a b-6 b c+4 c a\right)$
$=8 a^3-27 b^3-8 c^3-36 a b c$
View full question & answer→Question 363 Marks
Evaluate the following using identities: $117 \times 83$
AnswerWe have, $117 \times 83=(100+17)(100-17)$
$=(100)^2-(17)^2\left[(a+b)(a-b)=a^2-b^2\right]$
$=10000-289[\text { Where } a=100 \text { and } b=17]$
$=9711$
Therefore, $117 \times 83 = 9711$
View full question & answer→Question 373 Marks
Simplify the following products:
$(2\text{x}^4 -4\text{x}^2+1)(2\text{x}^4-4\text{x}^2-1)$
Answer In the given problem, we have to find product of
$(2\text{x}^4 -4\text{x}^2+1)(2\text{x}^4-4\text{x}^2-1)$
On rearranging we get $\Big(\big[2\text{x}^4-4\text{x}^2\big]+1\Big)\Big(\big[2\text{x}^4-4\text{x}^2\big]-1\Big)$
We shall use the identity $(\text{x}-\text{y})(\text{x}+\text{y)}=\text{x}^2-\text{y}^2$
$\big(2\text{x}^4-4\text{x}^2+1\big)\big(2\text{x}^4-4\text{x}^2-1\big)\\=\big[2\text{x}^4-4\text{x}^2\big]^2-1^2$
$=\big[4\text{x}^8+16\text{x}^4-2\times2\text{x}^4\times4\text{x}^2-1\big]$
$=4\text{x}^8+16\text{x}^4-16\text{x}^6-1$
Hence the value of $(2\text{x}^4 -4\text{x}^2+1)(2\text{x}^4-4\text{x}^2-1)$ is $4\text{x}^8+16\text{x}^4-16\text{x}^6-1$
View full question & answer→Question 383 Marks
If $x = 3$ and $y = -1$, find the values of the following using in identity: $\Big(\frac{\text{x}}{4}-\frac{\text{y}}{3}\Big)\Big(\frac{\text{x}^2}{16}+\frac{\text{xy}}{12}+\frac{\text{y}^2}{9}\Big)$
AnswerWe have,$\Big(\frac{\text{x}}{4}-\frac{\text{y}}{3}\Big)\Big(\frac{\text{x}^2}{16}+\frac{\text{xy}}{12}+\frac{\text{y}^2}{9}\Big)$
$=\Big(\frac{\text{x}}{4}-\frac{\text{y}}{3}\Big)\bigg[\Big(\frac{\text{x}}{4}\Big)^2+\frac{\text{x}}{4}\times\frac{\text{y}}{3}+\Big(\frac{\text{y}}{3}\Big)^2\bigg]$
$=\Big(\frac{\text{x}}{4}\Big)^3-\Big(\frac{\text{y}}{3}\Big)^3$ $\big[\because\text{a}^3-\text{b}^3=(\text{a}-\text{b})(\text{a}^2+\text{ab}+\text{b}^2)\big]$
$=\frac{\text{x}^3}{64}-\frac{\text{y}^3}{27}$
$=\frac{(3)^3}{64}-\frac{(-1)^3}{27}$ $\big[\because\text{x}=3,\ \text{y}=-1\big]$
$=\frac{27}{64}+\frac{1}{27}$
$=\frac{729+64}{1728}=\frac{793}{1728}$
$\therefore\Big(\frac{\text{x}}{4}-\frac{\text{y}}{3}\Big)\Big(\frac{\text{x}^2}{16}+\frac{\text{xy}}{12}+\frac{\text{y}^2}{9}\Big)=\frac{793}{1728}$
View full question & answer→Question 393 Marks
If $a + b + c = 0$, then write the value of $\frac{\text{a}^2}{\text{bc}}+\frac{\text{b}^2}{\text{ca}}+\frac{\text{c}^2}{\text{ab}}.$
AnswerWe have to find the value of $\frac{\text{a}^2}{\text{bc}}+\frac{\text{b}^2}{\text{ca}}+\frac{\text{c}^2}{\text{ab}}$
Given $a + b + c = 0$
Using identity $a^3 + b^3 + c^3 - 3abc = (a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca)$
Put $a + b + c = 0 a^3 + b^3 + c^3 - 3abc = (0)(a^2 + b^2 + c^2 - ab - bc - ca) a^3 + b^3 + c^3 - 3$
$abc$ $= 0 a^3 + b^3 + c^3 = 3abc$
$\frac{\text{a}^3}{\text{abc}}+\frac{\text{b}^3}{\text{abc}}+\frac{\text{c}^3}{\text{abc}}=3$
$\frac{\text{a}\times\text{a}\times\text{a}}{\text{abc}}+\frac{\text{b}\times\text{b}\times\text{b}}{\text{abc}}+\frac{\text{c}\times\text{c}\times\text{c}}{\text{abc}}=3$
$\frac{\text{a}^2}{\text{bc}}+\frac{\text{b}^2}{\text{ca}}+\frac{\text{c}^2}{\text{ab}}=3$
Hence the value of $\frac{\text{a}^2}{\text{bc}}+\frac{\text{b}^2}{\text{ca}}+\frac{\text{c}^2}{\text{ab}}$ is $3$.
View full question & answer→Question 403 Marks
Evaluate the following:$ (99)^3$
AnswerWe know that $(a-b)^3=a^3-b^3-3 a b(a-b)$
$\Rightarrow(99)^3$ can be written as $(100-1)^3$
Here, $a =100$ and $b =1(99)^3$
$= (100 - 1)^3 = (100)^3 - (1)^3 - 3(100)(1)(100 - 1)$
$= 1000000 - 1 - (300 \times 99)$
$= 1000000 - 1 - 29700$
$= 1000000 - 29701 = 970299$
The value of $(99)^3=970299$
View full question & answer→Question 413 Marks
If $x = -2$ and $y = 1$, by using an identity find the value of the following:
$\Big(\frac{2}{\text{x}}-\frac{\text{x}}{2}\Big)\Big(\frac{4}{\text{x}^2}+\frac{\text{x}^2}{4}+1\Big)$
AnswerWe have,
$\Big(\frac{2}{\text{x}}-\frac{\text{x}}{2}\Big)\Big(\frac{4}{\text{x}^2}+\frac{\text{x}^2}{4}+1\Big)$
$=\Big(\frac{2}{\text{x}}-\frac{\text{x}}{2}\Big)\bigg[\Big(\frac{2}{\text{x}}\Big)^2+\Big(\frac{\text{x}}{2}\Big)+\frac{2}{\text{x}}\times\frac{\text{x}}{2}\bigg]$
$=\Big(\frac{2}{\text{x}}\Big)^3-\Big(\frac{\text{x}}{2}\Big)^3$ $\big[\therefore\text{a}^3-\text{b}^3=(\text{a}-\text{b})(\text{a}^2+\text{b}^2+2\text{ab})\big]$
$=\frac{8}{\text{x}^3}-\frac{\text{x}^3}{8}$
$=\frac{8}{(-2)^3}-\frac{(-2)^3}{8}$ $\big[\therefore\text{x}=-2\big]$
$=\frac{8}{-8}+\frac{8}{8}$
$=-1+1=0$
$\therefore\Big(\frac{2}{\text{x}}-\frac{\text{x}}{2}\Big)\Big(\frac{4}{\text{x}^2}+\frac{\text{x}^2}{4}+1\Big)=0$
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