Question 1: Expand / Simplify the following:

(i) $(3x+4y)^2$

$(3x+4y)^2 = (3x)^2+2 \times 3x \times 4y + (4y)^2 = 9x^2 + 24xy + 16y^2$

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(ii) $(\sqrt{2}x-3y)^2$

$(\sqrt{2}x-3y)^2 = (\sqrt{2}x)^2 + 2 \times (\sqrt{2}x) \times (-3y) + (-3y)^2 = 2x^2 - 6\sqrt{2}xy + 9y^2$

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(iii) $\Big( 2x-$ $\frac{1}{3x}$ $\Big)^2$

$\Big( 2x-$ $\frac{1}{3x}$ $\Big)^2 = (2x)^2 + 2 \times (2x) \times ( -$ $\frac{1}{3x}$ $) + ($ $\frac{1}{3x})^2$ $= 4x^2-$ $\frac{4}{3}$ $+$ $\frac{1}{9x^2}$

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(iv) $(x-1)(x+1)(x^2+1)(x^4+1)$

$(x-1)(x+1)(x^2+1)(x^4+1)$

$= (x^2-1)(x^2+1)(x^4+1)$

$= (x^4-1)(x^4+1)$

$= (x^8-1)$

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(v) $\Big(x-$ $\frac{1}{x}$ $\Big) \Big(x+$ $\frac{1}{x}$ $\Big) \Big( x^2+$ $\frac{1}{x^2}$ $\Big) \Big( x^4+$ $\frac{1}{x^4}$ $\Big)$

$\Big(x-$ $\frac{1}{x}$ $\Big) \Big(x+$ $\frac{1}{x}$ $\Big) \Big( x^2+$ $\frac{1}{x^2}$ $\Big) \Big( x^4+$ $\frac{1}{x^4}$ $\Big)$

$=$ $\Big(x^2-$ $\frac{1}{x^2}$ $\Big) \Big( x^2+$ $\frac{1}{x^2}$ $\Big) \Big( x^4+$ $\frac{1}{x^4}$ $\Big)$

$=$ $\Big(x^4-$ $\frac{1}{x^4}$ $\Big) \Big( x^4+$ $\frac{1}{x^4}$ $\Big)$

$=$ $\Big(x^8-$ $\frac{1}{x^8}$ $\Big)$

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(vi) $(2x-$ $\frac{3}{x}$ $+1)(2x+$ $\frac{3}{x}$ $+ 1)$

$(2x-$ $\frac{3}{x}$ $+1)(2x+$ $\frac{3}{x}$ $+ 1)$

$= (2x+1 -$ $\frac{3}{x}$ $)(2x+1 +$ $\frac{3}{x}$ $)$

$= (2x+1)^2 - \Big($ $\frac{3}{x}$ $\Big)^2$

$= 4x^2 + 4x + 1 -$ $\frac{9}{x^2}$

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(vii) $(2x+5y+3)(2x+5y+4)$

Let $2x+5 = a$

$\Rightarrow (2x+5y+3)(2x+5y+4)$

$= (a+3)(a+4)$

$= a^2 + 7a + 12$

$= (2x+5)^2 + 7(2x+5)+ 12$

$= 4x^2 + 20x + 25 + 14x + 35 + 12$

$= 4x^2 + 34x + 72$

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(viii)  $(1.5x^2-0.3y^2)(1.5x^2+0.3y^2)$

$(1.5x^2-0.3y^2)(1.5x^2+0.3y^2) = (1.5x^2)^2 - (0.3y^2)^2 = 2.25x^4-0.09y^4$

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(ix) $(x^3-3x^2-x)(x^2-3x+1)$

$(x^3-3x^2-x)(x^2-3x+1)$

$= x(x^2+3x+1)(x^2-3x+1)$

$= x \Big( (x^2+3x)^2 - 1 \Big)$

$= x(x^4-6x^3+9x^2-1)$

$= x^5-6x^4+9x^3-1$

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(x) $(2x^4-4x^2+1)(2x^4-4x^2-1)$

$(2x^4-4x^2+1)(2x^4-4x^2-1)$

$= (2x^4-4x^2)^2 - 1$

$= 4x^8-16x^6+16x^4-1$

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(xi) $\Big(x+$ $\frac{2}{x}$ $-3\Big) \Big(x-$ $\frac{2}{x}$ $-3 \Big)$

$\Big(x+$ $\frac{2}{x}$ $-3\Big) \Big(x-$ $\frac{2}{x}$ $-3 \Big)$

$= \Big( (x - 3 )+$ $\frac{2}{x}$ $\Big) \Big( (x-3) -$ $\frac{2}{x}$ $\Big)$

$= x^2-6x+9 -$ $\frac{4}{x^2}$

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(xii) $(5-2x)(5+2x)(25+4x^2)$

$(5-2x)(5+2x)(25+4x^2)$ $= (25-4x^2)(25+4x^2)$ $= 625-16x^4$

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(xiii) $(x+2y+3)(x+2y+7)$

$(x+2y+3)(x+2y+7)$

$= (x+2y)^2 +10(x+2y) + 21$

$= x^2+4xy+4y^2+10x+20y+21$

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(xiv) $(x+1)(x+2)(x+3)$

$(x+1)(x+2)(x+3)$

$= (x^2+3x+2)(x+3)$

$= x^3+3x^2+2x+3x^2+9x+6$

$= x^3 +6x^2 +11x +6$

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(xv) $(a+2b+c)^2$

Note: We will use the following identify: ${(a+b+c)}^2=\left(a^2+b^2+c^2\right)+2(ab+bc+ca)$

$(a+2b+c)^2$ $= a^2 + 4b^2 +c^2 + 4ab + 4bc+ 2ca$

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(xvi) $\Big($ $\frac{x}{y}$ $+$ $\frac{y}{z}$ $+$ $\frac{z}{x}$ $\Big)^2$

$\Big($ $\frac{x}{y}$ $+$ $\frac{y}{z}$ $+$ $\frac{z}{x}$ $\Big)^2$

$= (x^4+y^4+z^4 + 2x^2y^2-2y^2z^2-2z^2x^2) - (x^4+y^4+z^4 - 2x^2y^2-2y^2z^2+2z^2x^2)$

$= x^4+y^4+z^4 + 2x^2y^2-2y^2z^2-2z^2x^2 -x^4-y^4-z^4 + 2x^2y^2+2y^2z^2-2z^2x^2$

$= 4x^2y^2 - 4z^2x^2$

$= 4x^2 (y^2-z^2)$

$= 4x^2(y+z)(y-z)$

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(xvii) $(-2x+3y+2z)^2$

$(-2x+3y+2z)^2$ $= 4x^2+9y^2+4z^2 -12xy+12yz-8zx$

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(xviii) $(a+b+c)^2 - (a-b+c)^2$

$(a+b+c)^2 - (a-b+c)^2$

$= \Big( a+b+c + a-b+c \Big) \Big( a+b+c -a +b -c \Big)$

$= ( 2a+2c ) ( 2b ) = 4ab + 4bc$

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(xix) $(x^2+y^2+z^2)^2-(x^2-y^2+z^2)^2$

$(x^2+y^2+z^2)^2-(x^2-y^2+z^2)^2$

$= \Big( x^2+y^2+z^2 + x^2-y^2+z^2 \Big) \Big( x^2+y^2+z^2 - x^2 +y^2 - z^2 \Big)$

$= (2x^2+2z^2)(2y^2) = 4x^2y^2+4y^2z^2$

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(xx) $(x+y+z)^2+$ $\Big( x+$ $\frac{y}{2}$ $+$ $\frac{z}{3}$ $\Big)^2 - \Big($ $\frac{x}{2}$ $+$ $\frac{y}{3}$ $+$ $\frac{z}{4}$ $\Big)^2$

$(x+y+z)^2+$ $\Big( x+$ $\frac{y}{2}$ $+$ $\frac{z}{3}$ $\Big)^2 - \Big($ $\frac{x}{2}$ $+$ $\frac{y}{3}$ $+$ $\frac{z}{4}$ $\Big)^2$

$= x^2+y^2+z^2 + 2xy + 2yz+2zx + x^2 +$ $\frac{1}{4}$ $y^2 +$ $\frac{1}{9}$ $z^2 + xy +$ $\frac{1}{3}$ $yz +$

$\frac{2}{3}$ $zx -$ $\frac{1}{4}$ $x^2 -$ $\frac{1}{9}$ $y^2 -$ $\frac{1}{16}$ $z^2 -$ $\frac{1}{3}$ $xy -$ $\frac{1}{6}$ $yz -$ $\frac{1}{4}$ $zx$

$= x^2( 1 + 1 -$ $\frac{1}{4}$ $) + y^2(1+$ $\frac{1}{4}$ $-$ $\frac{1}{9}$ $) +z^2(1 +$ $\frac{1}{9} -\frac{1}{16}$ $)$  $\\ +xy(2+1-$ $\frac{1}{3}$ $)$

$+ yz(2 +$ $\frac{1}{3}$ $-$ $\frac{1}{6}$ $)$ $+zx(2+$ $\frac{2}{3}$ $-$ $\frac{1}{4}$ $)$

$=$ $\frac{7}{4}$ $x^2$ $+$ $\frac{41}{36}$ $y^2$ $+$ $\frac{151}{144}$ $z^2$ $+$ $\frac{8}{3}$ $xy$ $+$ $\frac{13}{6}$ $yz$ $+$ $\frac{29}{12}$ $zx$

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(xx) $(x^2-x+1)^2-(x^2+x+1)^2$

$(x^2-x+1)^2-(x^2+x+1)^2$

$= (x^2-x+1 + x^2+x+1)(x^2-x+1 - x^2-x-1)$

$= (2x^2+2)(-2x) = -4x^3-4x$

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(xxi) $\Big( x+$ $\frac{2}{x}$ $\Big)^2 + \Big( x-$ $\frac{2}{x}$ $\Big)^2$

We will use the identity ${a^3+b^3= \ (a+b)}^3-3ab(a+b)$

$\Rightarrow a = x+$ $\frac{2}{x}$

$\Rightarrow b = x-$ $\frac{2}{x}$

$\Rightarrow a + b = 2x$

$\Big( x+$ $\frac{2}{x}$ $\Big)^2 + \Big( x-$ $\frac{2}{x}$ $\Big)^2$

$= (2x)^3 - 3(x +$ $\frac{2}{x}$ $)( x-$ $\frac{2}{x}$ $) (2x)$

$= 8x^3-6x^3+$ $\frac{24}{x}$ $= 2x^3 +$ $\frac{24}{x}$

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(xxii) $(2x-5y)^3-(2x+5y)^3$

We will use the identity ${a^3-b^3=\ (a-b)}^3+3ab(a-b)$

$\Rightarrow a = 2x-5y \ and \ b = 2x+5y$

$\Rightarrow a - b = 2x-5y-2x-5y = -10y$

Therefore $(2x-5y)^3-(2x+5y)^3$

$= (-10y)^3 + 3 (2x-5y)(2x+5y)(-10y)$

$= -1000y^3 -120yx^2+750y^3$

$= -250y^3 -120yx^2$

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(xxiii) $(4x-5y)(16x^2+20xy+25y^2)$

We will use the identity $a^3-b^3 = (a-b)(a^2 +ab + b^2)$

$(4x-5y)(16x^2+20xy+25y^2)$

$= (4x-5y)\Big[ (4x)^2 + (4x)(5y) + (5y)^2 \Big]$

$= (4x)^3 -(5y)^3 = 64x^3 - 125y^3$

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(xxiv) $(x^3+1)(x^6-x^3+1)$

We will use the identity $a^3-b^3 = (a-b)(a^2 +ab + b^2)$

$(x^3+1)(x^6-x^3+1)$

$= (x^3+1) \Big[ (x^3)^2 -(x^3)(1) +(1)^2 \Big]$

$= (x^3)^3 +(1)^3 = x^9 +1$

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(xxv) $(4x-3y+2z)(16x^2+9y^2+4z^2+12xy+6yz-8zx)$

We will use the identity $a^3+b^3+c^3 -3abc = (a+b+c)(a^2+b^2+c^2-ab-bc-ca)$

$(4x-3y+2z)(16x^2+9y^2+4z^2+12xy+6yz-8zx)$

$= (4x-3y+2z) \Big[ (4x)^2 +(-3y)^2 + (2z)^2 - (4x)(-3y) - (-3y)(2z) - (2z)(4x) \Big]$

$= (4x)^3+(-3y)^3+(2z)^3 - 3(4x)(-3y)(2z)$

$= 64x^3 - 27y^3+8z^3 +72xyz$

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Question 2: Evaluate the following identities:

(i) $103 \times 97$

$103 \times 97 = (100+3)(100-3) = 100^2 - 3^2 = 10000-9 = 9991$

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(ii) $(97)^2$

$(97)^2 = (100-3)^2 = 10000 -600 +9 = 9409$

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(iii) $0.54 \times 0.54 -0.46 \times 0.46$

$0.54 \times 0.54 -0.46 \times 0.46 = 0.54^2-0.46^2 = (0.54+0.46)(0.54-0.46) = 1 \times 0.08 = 0.08$

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(iv) $(0.98)^2$

$(0.98)^2 = (1-0.02)^2 = 1 -0.04+0.0004 = 0.9604$

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(v) $991 \times 1009$

$991 \times 1009 = (1000 -9) (1000 + 9) = 1000^2 -9^2 = 1000000-81 = 999919$

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(vi) $(1002)^3$

We will use the identity ${(a+b)}^3=\ a^3+b^3+3ab(a+b)$

$(1002)^3 = (1000+2)^3 = 1000^3 + 2^3 + 3\times 1000 \times 2 (1000 + 2)$

$= 1000000000+8 + 6012000= 1006012008$

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(vii) $(999)^3$

We will use the identity ${(a-b)}^3=\ a^3-b^3-3ab(a-b)$

$(999)^3 = (1000 - 1)^3 = 1000^3 -1^3-3 \times 1000 \times 1 (1000-1)$

$= 1000000000-1-299700 = 997002999$

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(viii) $(103)^3$

We will use the identity ${(a+b)}^3=\ a^3+b^3+3ab(a+b)$

$(103)^3 = (100 + 3)^3 = 100^3 + 3^3 + 3 \times 100 \times 3 (100+3)$

$= 1000000+27+92700 = 1092727$

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(ix) $(10.4)^3$

$(10.4)^3 = (10+0.4)^3 = 10^3 +0.4^3 + 3 \times 10 \times 0.4 (10+0.4)$

$= 1000 + 0.064+124.8 = 1124.864$

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(x) $(99)^3$

We will use the identity ${(a-b)}^3=\ a^3-b^3-3ab(a-b)$

$(99)^3 = (100 -1 )^3 = 100^3 - 1^3 - 3 \times 100 \times 1 (100 -1 )$

$= 1000000-1-29700=970299$

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(xi) $111^3 - 89^3$

We will use the identity  ${a^3-b^3=\ (a-b)}^3+3ab(a-b)$

$111^3 - 89^3 = (100+11)^3 - (100-11)^3$

$= (100+11-100+11)^3 + 3 (100+11)(100-11) (22)$

$=22^3 + 3 \times 111 \times 89 \times 22$

$= 662662$

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(xii) $104^3+96^3$

We will use the identity  $a^3+b^3= (a+b)^3-3ab(a+b)$

$104^3+96^3$

$= (100+4)^3 + (100-4)^3$

$= (100+4+100-4) - \Big[ (100+4+100-4)^3 - 3 (100+4)(100-4)(100+4+100-4) \Big]$

$= 200^3 - 3 \times 104 \times 96 \times 200$

$= 2009600$

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(xiii) $\Big($ $\frac{1}{2}$ $\Big)^3$ $+$ $\Big($ $\frac{1}{3}$ $\Big)^3$ $-$ $\Big($ $\frac{5}{6}$ $\Big)^3$

We are going to use this identity $a^3+b^3+c^3 -3abc = (a+b+c)(a^2+b^2+c^2-ab-bc-ca)$

$\Big($ $\frac{1}{2}$ $\Big)^3$ $+$ $\Big($ $\frac{1}{3}$ $\Big)^3$ $-$ $\Big($ $\frac{5}{6}$ $\Big)^3$

$=$ $($ $\frac{1}{2}$ $+$ $\frac{1}{3}$ $-$ $\frac{5}{6}$ $) \Big ( ($ $\frac{1}{2})^2$ $+$ $($ $\frac{1}{3}$ $)^2$ $+$ $(-$ $\frac{5}{6}$ $)^2$ $-$ $\frac{1}{2}$ $\times$ $\frac{1}{3}$ $+$ $\frac{1}{3}$ $\times$ $\frac{5}{6}$ $+$ $\frac{5}{6}$ $\times$ $\frac{1}{2}$ $\Big)$ $+ 3 \times$ $\frac{1}{2}$ $\times$ $\frac{1}{3}$ $\times (-$ $\frac{5}{6}$ $)$

$=$ $\frac{0}{6}$ $\Big ( ($ $\frac{1}{2})^2$ $+($ $\frac{1}{3}$ $)^2 + (-$ $\frac{5}{6})^2$ $-$ $\frac{1}{2}$ $\times$ $\frac{1}{3}$ $+$ $\frac{1}{3}$ $\times$ $\frac{5}{6}$ $+$ $\frac{5}{6}$ $\times$ $\frac{1}{2} \Big)$ $-$ $\frac{5}{12}$

$=$ $\frac{-5}{12}$

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Question 3:

(i) If the number $a$ is $7$ more than number $b$ and the sum of the squares of $a$ and $b$ is $85$, find the product of $ab$

$a+b = 7 \Rightarrow a - b = 7$

$a^2 + b^2 = 85$

Therefore $(a-b)^2 + 2ab = 85$

$\Rightarrow 7^2 + 2ab = 85$

$\Rightarrow 2ab = 85 - 49 = 36$

$\Rightarrow ab = 18$

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(ii) If the number $x$ is $3$ less than the number $y$ and the sum if the square of $x$ and $y$ is $29$, find $xy$

$x+3 = 7 \Rightarrow x - y = -3$

$x^2 + y^2 = 29$

$(x-y)^2 +2xy = 29$

$\Rightarrow (-3)^2 + 2xy = 29$

$\Rightarrow 2xy = 29 - 9 = 20$

$\Rightarrow xy = 10$

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(iii) If the sum of two numbers is $7$ and the sum of their cubes is $133$, find the sum of their squares.

$a+ b = 7$

$a^3 + b^3 = 133$

$(a+b)^3 = a^3 +b^3 +3ab(a+b)$

$7^2 = 133 + 3ab (7)$

$21ab = 133 - 49 = 84$

$\Rightarrow ab = 4$

Therefore $(a+b)^2 -2ab = a^2 + b^2$

$\Rightarrow 7^2 - 8 = a^2 +b^2$

$\Rightarrow a^2 +b^2 = 41$

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Exercise 5.1 continued…