Question 13 Marks
Evaluate the following: $(98)^3$
AnswerWe know that $(a - b)^3 = a^3 - b^3 - 3ab(a - b)$
$\Rightarrow (98)^3$ can be written as $(100 - 2)^3$
Here$, a = 100$ and $b = 2$
$(98)^3 = (100 - 2)^3$
$= (100)^3 - (2)^3 - 3(100)(2)(100 - 2)$
$= 1000000 - 8 - (600 \times 102)$
$= 1000000 - 8 - 58800$
$= 941192$
The value of $(98)^3$
$ = 941192$
View full question & answer→Question 23 Marks
If $9x^{2 }+ 25y^{2 }= 181$ and $xy = -6,$ find the value of $3x + 5y.$
AnswerWe have,
$(3x + 5y)^2 = (3x)^2 + (5y)^2 + 2 \times 3x \times 5y$
$\Rightarrow (3x + 5y)^2$
$ = 9\times ^2 + 25y^2 + 30xy$
$= 181 + 30(-6) [$Since$, 9x^{2 }+ 25y^{2 }= 181$ and $xy = - 6]$
$\Rightarrow (3x + 5y)^2 = 1$
$\Rightarrow (3x + 5y)^2 = (\pm 1)2$
$\Rightarrow 3x + 5y = \pm1$
View full question & answer→Question 33 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 43 Marks
If $3x - 2y = 11$ and $xy = 12,$ find the value of $27x^3 - 8y^3.$
AnswerGiven,$ 3x - 2y = 11, xy = 12$
We know that $(a – b)^3 = a^3 - b^3 - 3ab(a + b)$
$(3x - 2y)^3 = 11^3$
$\Rightarrow 27x^3 - 8y^{3 }- (18 \times 12 \times 11) = 1331$
$\Rightarrow 27x^3 - 8y^{3 }- 2376 = 1331$
$\Rightarrow 27x^3 - 8y^{3 }= 1331 + 2376$
$\Rightarrow 27x^3 - 8y^{3 }= 3707$
Hence, the value of $27x^3 - 8y^{3 }= 3707$
View full question & answer→Question 53 Marks
If $a - b = 4$ and $ab = 21,$ find the value of $a^3 - b^3.$
AnswerIn the given problem, we have to find the value of $a^3 - b^3$
Given $a - b = 4, ab = 21$
We shall use the identity $(a - b)^3 = a^3 - b^3 - 3ab(a - b)$
Here putting $a - b = 4, ab = 21,$
$\Rightarrow (4)^3 = a^3 - b^3 + 3(21)(4)$
$\Rightarrow 64 = a^3 - b^3 - 252$
$\Rightarrow 64 – 252 = a^3 - b^3$
$\Rightarrow 316 = a^{3 }- b^3$
Hence, the value of $a^{3 }- b^3$ is $316.$
View full question & answer→Question 63 Marks
Evaluate the following using identities : $991 \times 1009$
AnswerWe have,
$991 \times 1009$
$= (1000 - 9)(1000 + 9)$
$= (1000)^2 - (9) [(a + b)(a - b) = a^2 - b^2]$
$= 1000000 - 81 [$Where $a = 1000$ and $b = 9]$
$= 999919$
Therefore$, 991 \times 1009 = 999919$
View full question & answer→Question 73 Marks
If $a + b + c = 0$ and $a^2 + b^2 + c^2 = 16,$ find the value of $ab + bc + ca.$
AnswerWe know that,
$\big[\therefore(a + b + c)^2 = a^{2 }+ b^{2 }+ c^2 + 2ab + 2bc + 2ca\big]$
$(0)^2 = 16 + 2(ab + bc + ca)$
$2(ab + bc + ca)= -16$
$ab + bc + ca = -8$
Hence, value of required express $ab + bc + ca = -8$
View full question & answer→Question 83 Marks
If x = 3 and y = -1, find the values of the following using in identity:
$\Big(\frac{5}{\text{x}}+5\text{x}\Big)\Big(\frac{25}{\text{x}^2}-25+25\text{x}^2\Big)$
Answer We 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 93 Marks
If $a + b = 10$ and $ab = 21,$ find the value of $a^3 + b^3.$
AnswerGiven,
$a + b = 10, ab = 21$
We know that,$ (a + b)^3 = a^3 + b^3 + 3ab(a + b) ...1$
Substitute $a + b = 10, ab = 21$ 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 103 Marks
Evaluate the following: $(9.9)^3$
AnswerWe know that $(a - b)^3 = a^3 - b^3 - 3ab(a - b)$
$\Rightarrow (9.9)^3$ can be written as $(10 - 0.1)^3$
Here$, a = 10$ and $b = 0.1$
$(9.9)^3 = (10 - 0.1)^3$
$= (10)^3 - (0.1)^3 - 3(10)(0.1)(10 - 0.1)$
$= 1000 - 0.001 - (3 \times 9.9)$
$= 1000 - 0.001 - 29.7$
$= 1000 - 29.701$
$= 970.299$
The value of $(9.9)^3 = 970.299$
View full question & answer→Question 113 Marks
Evaluate the following : $(103)^3$
AnswerWe know that $(a + b)^3 = a^3 + b^3 + 3ab(a + b)$
$\Rightarrow (103)^3$ can be written as $(100 + 3)^3$
Here$, a = 100$ 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 123 Marks
If$ x = -2$ and $y = 1,$ by using an identity find the value of the following : $(4y^2 - 9x^2)(16y^4 + 36x^2y^2 + 81x^4)$
AnswerWe have,
$(4y^2 - 9x^2)(16y^4 + 36x^2y^2 + 81x^4)$
$= (4y^2 - 9x^2)\big[(4y^2)^2 + 4y^2 \times 8x^2 + (9x^2)^2\big]$
$= (4y^2)^3 - (9x^2)^3\big[\because a^3 - b^3 = (a - b)(a^2 + ab + b^2)\big]$
$= 64y^6 - 729x^6$
$= 64(1)^6 - 729(-2)^6 \big[\because x = 2$ and $y = 1\big]$
$= 64 - 729 \times 64$
$= 64 - 46656$
$= -465992$
$\therefore (4y^2 - 9x^2)(16y^4 + 36x^2y^2 + 81x^4)$
$ = -465992$
View full question & answer→Question 133 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[a^3 + 3ab^2]$
here, $a= 100, b = 4$
$(100 + 4)^3 - (100 - 4)^{3 }$
$= 2[100^3 + 3(4)^2(100)]$
$= 2[1000000 + 4800]$
$= 2[1004800]$
$= 200960$
The value of $104^3 + 96^3 $
$= 2009600$
View full question & answer→Question 143 Marks
Find the following products : $(3x + 2y + 2z)(9x^2 + 4y^2 + 4z^2 - 6xy - 4yz - 6zx)$
AnswerWe have,
$(3x + 2y + 2z)(9x^2 + 4y^2 + 4z^2 - 6xy - 4yz - 6zx)$
$= (3x + 2y + 2z)\big((3x)^2 + (2y)^2 + (2z)^2 - 3x \times 2y \times 2z - 2z \times 3x\big)$
$= (3x)^3 + (2y)^3 + (2z)^3 - 3 \times 3x \times 2y \times 2z \big[\because a^3 + b^3 + c^3 $$= 3abc = (a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca)\big]$
$= 27x^3 + 8y^3 + 8z^3 - 36xyz$
$\therefore (3x + 2y + 2z)(9x^2 + 4y^2 + 4z^2 - 6xy - 4yz - 6zx)$
$ = 27x^3 + 8y^3 + 8z^3 - 36xyz$
View full question & answer→Question 153 Marks
Evaluate the following : $(598)^3$
AnswerWe know that $(a - b)^3 = a^3 - b^3 - 3ab(a - b)$
$\Rightarrow (598)^3$ can be written as $(600 - 2)^3$
Here, $a = 600$ and $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 163 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 173 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[b^3 + 3ba^2]$
here, $a = 93, b = 107$
$(100 - 7)^3 - (100 + 7)^{3 }= -2[7^3 + 3(100)^2(7)]$
$= -2[343 + 210000]$
$= -2[210343]$
$= -420686$
The value of $93^3 - 107^3 $
$= -420686$
View full question & answer→Question 183 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[b^3 + 3ab^2]$
here,$ a = 100 b = 11$
$(100 + 11)^3 - (100 - 11)^{3 }= 2[11^3 + 3(100)^2(11)]$
$= 2[1331 + 330000]$
$= 2[331331]$
$= 662662$
The value of $111^3 - 89^3 $
$= 662662$
View full question & answer→Question 193 Marks
Find the value of $4x^2 + y^2 + 25z^2 + 4xy - 10yz - 20zx$ 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 203 Marks
Evaluate the following: $(10.4)^3$
AnswerWe know that $(a + b)^3 = a^3 + b^3 + 3ab(a + b)$
$\Rightarrow (10.4)^3$ can be written as $(10 + 0.4)^3$
Here, $a = 10$ and $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 213 Marks
If $a + b + c = 9$ and $ab + bc + ca = 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 223 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[a^3 + 3ab^2]$
here,$ a = 40, b = 4$
$(40 + 6)^3 + (40 - 6)^{3 }= 2[40^3 + 3(6)^2(40)]$
$= 2[64000 + 4320]$
$= 2[68320]$
$= 1366340$
The value of $46^3 + 34^3 $
$= 1366340$
View full question & answer→Question 233 Marks
If $a^2 + b^2 + c^2 = 16$ and $ab + bc + ca = 10,$ find the value of $a + b + c.$
AnswerWe know that,
$(a + b + c)^2 = a^2 + b^2 + c^2 + 2(ab + bc + ca)$
$(x + y + z)^2 = 16 + 2(10)$
$(x + y + z)^2 = 36$
$\big(\text{x}+\text{y}+\text{z}\big) = \sqrt{36}$
$\big(\text{x}+\text{y}+\text{z}\big) =\pm6$
Hence, value of required expression $\big(\text{a}+\text{b}+\text{c}) =\pm6$
View full question & answer→Question 243 Marks
If $a + b = 7$ and $ab = 12,$ find the value of $a^2 + b^2$
AnswerWe have to find the value of $a^2 + b^2$
Given $a + b = 7, ab = 12$
Using identity $(a + b)^2 = a^2 + 2ab + b^2$
By substituting the value of $a + b = 7, ab = 12$ we get
$(a + b)^2 = a^2 + b^2 + 2 \times ab$
$(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 253 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 263 Marks
If $a + b = 8$ and $ab = 6,$ find the value of $a^3 + b^3.$
AnswerWe have,
$a^3 + b^3 = (a + b)(a^2 - ab + b^2)$
$= (a + b)(a^2 + b^2 - ab)$
$= (a + b)(a^2 + b^2 - ab + 2ab - 2ab) [$Adding and substracting $2ab$ in the second break$]$
$= (a + b)\big[(a^2 + b^2 + 2ab) - 3ab\big]$
$= (a + b)\big[(a + b)^2 - 3ab\big] \big[\because (a + b)^2 = a^2 + b^2 + 2ab\big]$
$= 8 \times \big[(8)^2 - 3 \times 6\big] \big[\because a + b = 8$ and $ab = 6\big]$
$= 8 \times [64 - 18]$
$= 8 \times 46$
$= 368$
$\therefore a^3 + b^3 = 368$
View full question & answer→Question 273 Marks
Evaluate the following using identities: $117 \times 83$
AnswerWe have,
$117 \times 83$
$= (100 + 17)(100 - 17)$
$= (100)^2 - (17)^{2 }[(a + b)(a - b) = a^2 - b^2]$
$= 10000 - 289 [$Where $a = 100$ and $b = 17]$
$= 9711$
Therefore,$ 117 \times 83 = 9711$
View full question & answer→Question 283 Marks
Simplify the following products:
$(2\text{x}^4 -4\text{x}^2+1)(2\text{x}^4-4\text{x}^2-1)$
AnswerIn 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 293 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 - 3abc = 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 303 Marks
Evaluate the following: $(99)^3$
AnswerWe know that$ (a - b)^3 = a^3 - b^3 - 3ab(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 313 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$
View full question & answer→Question 323 Marks
Find the following products: $(4x - 3y + 2z)(16x^2 + 9y^2 + 4z^2 + 12xy + 6yz - 8zx)$
AnswerWe have,
$(4x - 3y + 2z)(16x^2 + 9y^2 + 4z^2 + 12xy + 6yz - 8zx)$
$= (4x +(-3y) + 2z)\big[(4x)^2 + (-3y)^2 + (2z)^2 - (4x)(-3y) - (-3y)(2z) - (2z)(4x)\big]$
$= (4x)^3 + (-3y)^3 + (2z)^3 - 3 \times (4x) \times (-3y)(2z) \big[\because a^3 + b^3 + c^3 $$= 3abc = (a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca)\big]$
$= 64x^3 - 27y^3 + 8z^3 + 72xyz$
$\therefore (4x - 3y + 2z)(16x^2 + 9y^2 + 4z^2 + 12xy + 6yz - 8zx)$
$ = 64x^3 - 27y^3 + 8z^3 + 72xyz$
View full question & answer→Question 333 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 343 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 353 Marks
Simplify the following products:
$(\text{x}^3 -3\text{x}^2-\text{x})(\text{x}^2-3\text{x} + 1)$
AnswerIn 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 363 Marks
If $x = 3$ and $y = -1,$ find the values of the following using in identity : $(9y^2 - 4x^2)(81y^4 + 36x^2y^2 + 16x^4)$
AnswerWe have,
$(9y^2 - 4x^2)(81y^4 + 36x^2y^2 + 16x^4)$
$= (9y^2 - 4x^2)\big[(9y^2)^2 + 9y^2 \times 4x^2 + (4x^2)^2\big]$
$= (9y^2)^3 - (4x^2)^3\big[\because a^3 - b^3 = (a - b)(a^2 + ab + b^2)\big]$
$= 729y^6 - 64x^6$
$= 729 \times (-1)^6 - 64(3)^6 \big[\because x = 3$ and $y = -1\big]$
$= 729 - 64 \times 729$
$= 729 - 46656$
$= -45927$
$(9y^2 - 4x^2)(81y^4 + 36x^2y^2 + 16x^4)$
$ = -45927$
View full question & answer→Question 373 Marks
Find the following products: $(3x - 4y + 5z)(9x^2 + 16y^2 + 25z^2 + 12xy - 15zx + 20yz)$
AnswerWe have,
$(3x - 4y + 5z)(9x^2 + 16y^2 + 25z^2 + 12xy - 15zx + 20yz)$
$= (3x + (-4y) + 5z)\big((3x)2 + (-4y)2 + (5z)2 - (3x)(-4y) - (5z)(3x)\big)$
$= (3x)^3 + (-4y)^3 + (5z)^3 - 3(3x)(-4y)(5z)$
$\big[\because a^3 + b^3 + c^3 - 3abc = (a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca)\big]$
$= 27x^3 - 64y^3 + 125z^3 + 180xyz$
$\therefore (3x - 4y + 5z)(9x^2 + 16y^2 + 25z^2 + 12xy - 15zx + 20yz)$
$ = 27x^3 - 64y^3 + 125z^3 + 180xyz$
View full question & answer→Question 383 Marks
If $a - b = 6$ and $ab = 20,$ find the value of $a^3 - b^3.$
AnswerWe have,
$a^3 - b^3 = (a - b)(a^2 + ab + b^2)$
$= (a - b)(a^2 + ab + b^{2 }- 2ab + 2ab) [$Adding and substracting $2ab$ in the second break$]$
$= (a - b)\big[(a^2 + b^2 - 2ab) + 3ab\big]$
$= (a - b)\big[(a - b)^2 + 3ab\big] \big [\because (a - b)^2 = a^2 + b^2 - 2ab\big]$
$= 6 \times \big[(6)^2 + 3 \times 20\big] \big[\because a - b = 6$ and $ab = 20\big]$
$= 6 \times [36 + 60]$
$= 6 \times 96$
$= 576$
$\therefore a^3 - b^3 $
$= 576$
View full question & answer→Question 393 Marks
Find the following products: $(2a - 3b - 2c)(4a^2 + 9b^2 + 4c^2 + 6ab - 6bc + 4ca)$
AnswerWe have,
$(2a - 3b - 2c)(4a^2 + 9b^2 + 4c^2 + 6ab - 6bc + 4ca)$
$= (2a + (-3b) + (-2c)) + \big((2a)^2 + (-3b)^2 + (-2c)^2 - (2a)(-3b)(-2c) - (-2c)(2a)\big)$
$= (2a)^3 + (-3b)^3 + (-2c)^3 - 3(2a)(-3b)(-2c)$
$\big[\because a^3 + b^3 + c^3 - 3abc = (a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca)\big]$
$= 8a^3 - 27b^3 - 8c^3 - 36abc$
$\therefore (2a - 3b - 2c)(4a^2 + 9b^2 + 4c^2 + 6ab - 6bc + 4ca)$
$ = 8a^3 - 27b^3 - 8c^3 - 36abc$
View full question & answer→Question 403 Marks
If x = -2 and y = 1, by using an identity find the value of the following:
$\Big(5\text{y}+\frac{15}{\text{y}}\Big)\Big(25\text{y}^2-75+\frac{225}{\text{y}^2}\Big)$
Answer We 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 413 Marks
If x = 3 and y = -1, find the values of the following using in identity:
$\Big(\frac{\text{x}}{4}-\frac{\text{y}}{3}\Big)\Big(\frac{\text{x}^2}{16}+\frac{\text{xy}}{12}+\frac{\text{y}^2}{9}\Big)$
Answer We have,$\Big(\frac{\text{x}}{4}-\frac{\text{y}}{3}\Big)\Big(\frac{\text{x}^2}{16}+\frac{\text{xy}}{12}+\frac{\text{y}^2}{9}\Big)$
$=\Big(\frac{\text{x}}{4}-\frac{\text{y}}{3}\Big)\bigg[\Big(\frac{\text{x}}{4}\Big)^2+\frac{\text{x}}{4}\times\frac{\text{y}}{3}+\Big(\frac{\text{y}}{3}\Big)^2\bigg]$
$=\Big(\frac{\text{x}}{4}\Big)^3-\Big(\frac{\text{y}}{3}\Big)^3$ $\big[\because\text{a}^3-\text{b}^3=(\text{a}-\text{b})(\text{a}^2+\text{ab}+\text{b}^2)\big]$
$=\frac{\text{x}^3}{64}-\frac{\text{y}^3}{27}$
$=\frac{(3)^3}{64}-\frac{(-1)^3}{27}$ $\big[\because\text{x}=3,\ \text{y}=-1\big]$
$=\frac{27}{64}+\frac{1}{27}$
$=\frac{729+64}{1728}=\frac{793}{1728}$
$\therefore\Big(\frac{\text{x}}{4}-\frac{\text{y}}{3}\Big)\Big(\frac{\text{x}^2}{16}+\frac{\text{xy}}{12}+\frac{\text{y}^2}{9}\Big)=\frac{793}{1728}$
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