Question 13 Marks
If $x^2+\frac{1}{x^2}=18$; find $: x^3-\frac{1}{x^3}$
Answer$\left(x^3-\frac{1}{x}\right)^3 $
$ =x^3-\frac{1}{x^3}-3\left(x-\frac{1}{x}\right) $
$ \Rightarrow 64=x^3-\frac{1}{x^3}-3(4) $
$\Rightarrow x^3-\frac{1}{x^3} $
$=64+12$
$ =76 .$
View full question & answer→Question 23 Marks
If $x^2+\frac{1}{x^2}=18$; find : $x-\frac{1}{x}$
Answer$x^2+\frac{1}{x^2}=18 $
Using$\left(x-\frac{1}{x}\right)^2$
$ =x^2+\frac{1}{x^2}-2 $
$ \Rightarrow\left(x-\frac{1}{x}\right)^2$
$=18-2 $
$ =16$
$ \Rightarrow x-\frac{1}{x}$
$ =4 .$
View full question & answer→Question 33 Marks
Simplify:$(3a - 7b + 3)(3a - 7b + 5)$
Answer$(3a - 7b + 3)(3a - 7b + 5)$
$= 3a(3a - 7b + 5) -7b (3a - 7b + 5) + 3(3a - 7b + 5)$
$= 9a^2 - 21ab + 15a - 21ab + 49b^2 - 35b + 9a - 21b + 15$
$= 9a^2 - 42ab + 24a + 49b^2 - 56b + 15.$
View full question & answer→Question 43 Marks
Simplify:$(2x - 4y + 7)(2x + 4y + 7)$
Answer$(2x - 4y + 7)(2x + 4y + 7)$
$= (2x + 7 - 4y)(2x + 7 + 4y)$
$= (2x + 7)^2 - (4y)^2$
$= 4x^2+ 2(2x)(7) + 7^2 - 16y^2$
$= 4x^2 - 16y^2 + 28x + 49.$
View full question & answer→Question 53 Marks
Simplify:$(3x + 5y + 2z)(3x - 5y + 2z)$
Answer$(3x + 5y + 2z)(3x - 5y + 2z)$
$= (3x + 2z + 5y)(3x + 2z - 5y)$
$= (3x + 2z)^2 - (5y)^2$
$= 9x^2 + 2(3)(2z) + 4z^2 - 25y^2$
$= 9x^2 - 25y^2 + 4z^2 + 12xz.$
View full question & answer→Question 63 Marks
Simplify:$(x + 2y + 3z)(x^2 + 4y^2 + 9z^2 - 2xy - 6yz - 3zx)$
Answer$(x + 2y + 3z)(x^2 + 4y^2 + 9z^2 - 2xy - 6yz - 3zx)$
$= x(x^2 + 4y^2 + 9z^2 - 2xy - 6yz - 3zx) + 2y(x^2+ 4y^2 + 9z^2 - 2xy - 6yz - 3zx)$
$= x^3 + 4xy^2 + 9xz^2 - 2x^2y - 6xyz - 3zx^2 + 2x^2y $$+ 8y^3 + 18yz^2 - 4xy^2 - 12^2z - 6xyz + 3x^2z + 12y^2z + 27z^3 - 6xyz - 18yz^2 - 9xz^2$
$= x^3+ 8y^3 + 27z^3 - 18xyz.$
View full question & answer→Question 73 Marks
Simplify:$(2x + y)(4x^2 - 2xy + y^2)$
Answer$(2x + y)(4x^2 - 2xy + y^2)$
$= 2x(4x^2 - 2xy + y^2) + y(4x^2 - 2xy + y^2)$
$= 8x^3 - 4x^2y + 2xy^2 + 4x^2y - 2xy^2 + y2$
$= 8x^3 + y^3.$
View full question & answer→Question 83 Marks
Simplify:$(x + y - z)^2 + (x - y + z)^2$
Answer$(x + y - z)^2 + (x - y + z)^2$
$= x^2 + y^2 + z^2 + 2(x)(y) + 2(y)(-z) + 2(x)(-z) + x^2 + y^2 + z^2 + 2(x)(-y) + 2(-y) + 2(x)(z)$
$= x^2+ y^2 + z^2 + 2xy - 2yz - 2xz + x^2 + y^2 + z^2 - 2xy - 2yz + 2xz$
$= 2x^2 + 2y^2 + 2z^2- 4yz.$
View full question & answer→Question 93 Marks
Simplify:$\left(a-\frac{1}{a}\right)^2+\left(a+\frac{1}{a}\right)^2$
Answer$\left(a-\frac{1}{a}\right)^2+\left(a+\frac{1}{a}\right)^2$
$=(a)^2+\left(\frac{1}{a}\right)^2-2(a)\left(\frac{1}{a}\right)+(a)^2+\left(\frac{1}{a}\right)^2+2(a)\left(\frac{1}{a}\right) $
$ =a^2+\frac{1}{a^2}-2+a^2+\frac{1}{a^2}+2 $
$=2 a^2+\frac{2}{a^2} .$
View full question & answer→Question 103 Marks
Simplify:$(a + b)^3 + (a - b)^3$
Answer$(a + b)^3 + (a - b)^3$
$= a^3 + b^3+ 3ab (a + b) + a^3 - 3ab (a - b) - b^3$
$= a^3+ b^3 + 3a^2b + 3ab^2 + a^3 - 3a^2b + 3ab^2 - b^3$
$= 2a^3+ 6ab^2.$
View full question & answer→Question 113 Marks
Simplify:$(7a +5b)^2 - (7a - 5b)^2$
Answer$(7a + 5b)^2 - (7a - 5b)^2$
$= (7a)^2 + (5b)^2 + 2(7a)(5b) - [(7a)^2 + (5b)^2- 2(7a)(5b)]$
$= 49a^2+ 25b^2 + 70ab - [49a^2 + 25b^2 - 70ab]$
$= 70a + 70ab$
$= 140ab.$
View full question & answer→Question 123 Marks
If $x + 2y = 5,$ then show that $x^3 + 8y^3 + 30xy = 125.$
AnswerGiven $x + 2y = 5$
$(x + 2y)^3 = 53$
$\Rightarrow (x)^3 + (2y) + 3(x)(2y)(x + 2y) = 5^3 ....[$Using $(a + b)^3= (a)^3 + (b)^3 + 3ab (a + b)]$
$\Rightarrow (x)^3 + (2y)^3+ 6xy(x + 2y) = 125$
$\Rightarrow (x)^3+ (2y)^3+ 6xy(5) = 125$
$\Rightarrow x^3+ 8y^3+ 30xy = 125.$
View full question & answer→Question 133 Marks
If $2a - 3b = 10$ and $ab = 16;$ find the value of $8a^3 - 27b^3$.
Answer$2a - 3b = 10$
$(2a - 3b)^3 = (2a)^3 - (3b)^3 - 2(2a) (3b) (2a - 3b)$
$\Rightarrow 1000 = 8a^3 - 27b^3 - 12(16) (10)$
$\Rightarrow 8a^3 - 27b^3$
$= 1000 + 1920$
$= 2920.$
View full question & answer→Question 143 Marks
If $p - q = -1$ and $pq = -12,$ find $p^3 - q^3$
Answer$p - q = -1, pq = -12$
$(p - q)^3 = p^3- q^3 - 3pq (p - q)$
$\Rightarrow (-1)^3 = p^3 - q^3 - 3(-12) (-1)$
$\Rightarrow p^3 - q^3$
$= -14 + 36$
$= 35.$
View full question & answer→Question 153 Marks
If $x^3 + y^3 = 9$ and $x + y = 3$, find $xy.$
Answer$x^3 + y^3 = 9, x + y = 3$
$(x + y)^3 = x^3+ y^3+ 3xy (x + y)$
$\Rightarrow (3)^3 = 9 + 3xy (3)$
$\Rightarrow 27 = 9 + 9xy$
$\Rightarrow 9xy = 27 - 9 = 18$
$\Rightarrow xy = 2.$
View full question & answer→Question 163 Marks
Find the cube of: $4x + 7y$
AnswerUsing $(a + b)^3$
$= a^3 + b^3+ 3ab + (a + b)$
$(4x + 7y)^3$
$= (4x)^3+ (7y)^3 + 3(4x) (7y) (4x + 7y)$
$= 64x^3 + 343y^3 + 84xy (4x + 7y)$
$= 64x^3 + 343y^3 + 336x^2y + 588xy^2.$
View full question & answer→Question 173 Marks
If $r -\frac{1}{ r }=4 ;$ find $: r ^3-\frac{1}{ r ^3}$
Answer$\left(r-\frac{1}{r}\right)^3 $
$=r^3-\frac{1}{r^3}-3\left(r-\frac{1}{r}\right) $
$ \Rightarrow(4)^3=r^3-\frac{1}{r^3}-3(4) $
$=r^3-\frac{1}{r^3}$
$=64+12$
$ =76 .$
View full question & answer→Question 183 Marks
If $r -\frac{1}{ r }=4 ;$ find : $r ^4+\frac{1}{ r ^4}$
Answer$\left(r^2+\frac{1}{r^2}\right)^2$
$ =r^4+\frac{1}{\sqrt{4}}+2 $
$ \Rightarrow(18)^2=r^4+\frac{1}{r^4}+2 $
$\Rightarrow r^4+\frac{1}{r^4}$
$=324-2 $
$=322 .$
View full question & answer→Question 193 Marks
If $r -\frac{1}{ r }=4$; find: $r ^2+\frac{1}{ r ^2}$
Answer$\left(r-\frac{1}{r}\right)^2$
$=r^2+\frac{1}{r^2}-2$
$ \Rightarrow\left(4^2=r^2+\frac{1}{r^2}-2\right). $
$ =r^2+\frac{1}{r^2} $
$ =16+2 $
$=18 .$
View full question & answer→Question 203 Marks
Find the cube of: $2a - 5b$
AnswerUsing $(a - b)^3$
$= a^3 - b^3 - 3ab (a - b)$
$(2a - 5b)$
$= (2a)^3- (5b)^3 - 3(2a) (5b) (2a - 5b)$
$= 8a^3 - 125b^3 - 30ab (2a - 5b)$
$= 8a^3- 125b^3 - 60a^2b + 150ab^2.$
View full question & answer→Question 213 Marks
If $p +\frac{1}{ p }=6$; find : $p ^3+\frac{1}{ p ^3}$
Answer$\left( p +\frac{1}{ p }\right)^3 $
$ = p ^3+\frac{1}{ p ^3}+3\left( p +\frac{1}{ p }\right) $
$\Rightarrow 216= p ^3+\frac{1}{ p ^3}+3(6)$
$ = p ^3+\frac{1}{ p ^3} $
$=216-18 $
$ =198 .$
View full question & answer→Question 223 Marks
If $p+\frac{1}{p}=6 ;$ find $: p^4+\frac{1}{p^4}$
Answer$\left(p^2+\frac{1}{p^2}\right)$
$ =p^4+\frac{1}{p^4}+2$
$ =(34)^2$
$=p^4+\frac{1}{p^4}+2$
$=p^4+\frac{1}{p^4} $
$ =1158-2 $
$=1154 .$
View full question & answer→Question 233 Marks
If $p +\frac{1}{ p }=6_x^6$ find : $p ^2+\frac{1}{ p ^2}$
Answer$\left(p+\frac{1}{p}\right)^2$
$=p^2+\frac{1}{p^2}+2$
$\Rightarrow 36=p^2+\frac{1}{p^2}+2$
$=p^2+\frac{1}{p^2}$
$=36-2$
$=34 .$
View full question & answer→Question 243 Marks
If $x + y = 9, xy = 20$find: $x^2 - y^2.$
Answer$x + y = 9, xy = 20$
We know,
$(x - y) (x + y)$
$= x^2 - y^2$
$\Rightarrow x^2 - y^2$
$= (±1) (9)$
$= ±9.$
View full question & answer→Question 253 Marks
Evaluate, using $(a + b)(a - b)= a^2 - b^2,15.9 \times 16.1$
Answer$15.9 \times 16.1$
$= (16 - 0.1) \times (16 + 0.1)$
$= (16)^2 - (0.1)^2$
$= 256 - 0.01$
$= 255.99.$
View full question & answer→Question 263 Marks
Evaluate, using $(a + b)(a - b)= a^2 - b^2,4.9 \times 5.1$
Answer$4.9 \times 5.1$
$= (5 - 0.1) \times (5 + 0.1)$
$= (5)^2 - (0.1)^2$
$= 25 - 0.01$
$= 24.99.$
View full question & answer→Question 273 Marks
Evaluate, using $(a + b)(a - b)= a^2 - b^2,999 \times 1001$
Answer$999 \times 1001$
$= (1000 - 1) \times (1000 + 1)$
$= (1000)^2 - (1)^2$
$= 1000000 - 1$
$= 999999.$
View full question & answer→Question 283 Marks
Evaluate, using $(a + b)(a - b)= a^2 - b^2,399 \times 401$
Answer$399 \times 401$
$= (400 - 1) \times (400 + 1)$
$= (400)^2 - (1)^2$
$= 160000 - 1$
$= 159999.$
View full question & answer→Question 293 Marks
Evaluate the following without multiplying:$(1005)^2$
Answer$(1005)^2$
$= (1000 + 5)^2$
$= (1000)^2 + 2(1000)(5) + (5)^2$
$= 1000000 + 10000 + 25$
$= 1010025.$
View full question & answer→Question 303 Marks
Evaluate the following without multiplying:$(999)^2$
Answer$(999^{)2}$
$= (1000 - 1)^2$
$= (1000)^2 - 2(1000)(1) + (1)^2$
$= 1000000 - 2000 + 1$
$= 998001.$
View full question & answer→Question 313 Marks
Evaluate the following without multiplying:$(103)^2$
Answer$(103)^2$
$= (100 + 3)^2$
$= (100)^2 + 2(100)(3) + (3)^2$
$= 10000 + 600 + 9$
$= 10609.$
View full question & answer→Question 323 Marks
Evaluate the following without multiplying:$(95)^2$
AnswerUsing $(x + y)^2$
$= x^2+ 2xy + y^2,$
we get
$(95)^2$
$= (100 - 5)^2$
$= (100)^2 - 2(100)(5) + (5)^2$
$= 10000 - 1000 + 25$
$= 9025.$
View full question & answer→Question 333 Marks
Simplify by using formula :$(1 + a) (1 - a) (1 + a^2)$
Answer$(1 + a) (1 - a) (1 + a^2)$
$= [(1)^2 - (a)^2] (1 + a^2)$
$= (1 - a^2) (1 + a^2)$
$($Using identify $: (a + b)(a - b) = a^2 - b^2)$
$= (1)^2 - (a2)^2$
$= 1 - a^4.$
View full question & answer→Question 343 Marks
Simplify by using formula :$(a + b - c) (a - b + c)$
Answer$(a + b - c) (a - b + c)$
$= (a + b - c) [a - (b - c)]$
$= (a)^2 - (b - c)^2$
$($using identity $: (a + b) (a - b) = a^2 - b^2)$
$= a^2 - (b^2 + c^2 - 2bc)$
$= a^2- b^2- c^2 + 2bc.$
View full question & answer→Question 353 Marks
Find the squares of the following:$(2a + 3b - 4c)$
AnswerUsing $(a + b + c)^2$
$= a^2 + b^2+ c^2 + 2ab + 2bc + 2ac$
$(2a +3b - 4c)^2$
$= (2a)^2 + (3b)^2 + (4c)^2 + 2(2a)(3b) + 2(3b)(-4c) + 2(2a)(-4c)$
$= 4a^2 + 9b^2+ 16c^2 + 12ab - 24bc - 8ac.$
View full question & answer→Question 363 Marks
If $p^2 + q^2 + r^2 = 82$ and $pq + qr + pr = 18$; find $p + q + r.$
Answer$(p+q+r)^2$
$=p^2+q^2+r^2+2 p q+2 q r+2 p r$
$=8^2+2(18) $
$ =64+36$
$ =100$
$=p+q+r $
$ =\sqrt{100} $
$= \pm 10 .$
View full question & answer→Question 373 Marks
If $a^2 + b^2 + c^2 = 41$ and $a + b + c = 9;$ find $ab + bc + ca.$
Answer$(a + b + c)^2 = (9)^2$
$a^2 + b^2 + c^2 + 2ab + 2bc + 2ca = 81$
$\Rightarrow 41 + 2(ab + bc + ca) = 81$
$\Rightarrow 2(ab + bc + ca)$
$= 81 - 41$
$= 40$
$\Rightarrow ab + bc + ca$
$= 20.$
View full question & answer→Question 383 Marks
If $a ^2-7 a +1=0$ and $a =\neq 0$, find: $a ^2+\frac{1}{ a ^2}$
Answer$a+\frac{1}{a}=7 $
$ \Rightarrow a^2+\frac{1}{a^2}+2$
$ =49$
$ \Rightarrow a^2+\frac{1}{a^2} $
$ =49-2$
$=47 .$
View full question & answer→Question 393 Marks
If $x+\frac{1}{x}=3 ;$ find $x^2+\frac{1}{x^2}$
Answer$\left(x+\frac{1}{x}\right)^2 $
$ =x^2+\frac{1}{x^2}+2(x)\left(\frac{1}{x}\right) $
$ \Rightarrow(3)^2 $
$ =x^2+\frac{1}{x^2}+2$
$\Rightarrow x^2+\frac{1}{x^2}$
$=9-2 $
$ =7 .$
View full question & answer→