KC Sinha Mathematics Solution Class 9 Chapter 4 Algebraic identities exercise 4.5

Page 4.34

Exercise 4.5

Type 1
Problems based on factorization of algebraic expressions, expressible asthe sum or difference of two cubes.
WORKING RULE:
1. if in the given expression, any factor is common in each term , thentake out the common factors.
Now , use the following identities whichever is required
x³+y³=(x+y)(x²-xy+y²)
x³-y³=(x-y)(x²+xy+y²)
2. Replace y be in x³+y³ to get x³-y³

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Question 1

a³+27b³
Sol:
⇒(a)³+(3b)³
Using identity:
x³+y³=(x+y)(x²-xy+y²)
⇒(a+3b)[a²-(a)(3b)+(3b)²]
⇒(a+3b)(a²-3ab+9b²)

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Question 2

x³+125
Sol:
⇒x³+5³
Using identity:
x³+y³=(x+y)(x²-xy+y²)
⇒(x+5)[x²-(x)(5)+5²]
⇒(x+5)(x²-5x+25)

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Question 3

8a³+27b³
Sol:
⇒(2a)³+(3b)³
Using identity:
x³+y³=(x+y)(x²-xy+y²)
⇒(2a+3b)[(2a)²-(2a)(3b)+(3b)²]
⇒(2a+3b)(4a²+9b²-6ab)

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Question 4

x⁵+27x²
[Hint: x⁵+27x²=x²(x³+27)]
Sol:
[Taking common x²]
⇒x²(x³+27)
⇒x²(x³+3³)
Using identity:
x³+y³=(x+y)(x²-xy+y²)
⇒x²(x+3)(x²-3x+3²)
⇒x²(x+3)(x²-3x+9)

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Question 5

ab⁷+ba⁷
[Hint:ab⁷+ba⁷=ab(a⁶+b⁶)=ab[(a²)³+(b²)³]]
Sol:
[Taking common ab]
⇒ab(b⁶+a⁶)
⇒ab[(b²)³+(a²)³]
⇒ab[(a²)³+(b²)³]
Using identity:
x³+y³=(x+y)(x²-xy+y²)
⇒ab[(a²+b²)][(a²)²-(a²)(b²)+(b²)²]
⇒ab(a²+b²)(a⁴-a²b²+b⁴)

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Question 6

8(a+b)³ + 27(b+c)³
Sol:
[Hint: Given expression {2(a+b)}]³+{3(b+c)}³
={2(a+b)+3(b+c)}{4(a+b)²-6(a+b)(b+c)+9(b+c)²}
=(2a+5b+3c){(a+b)(4a+4b-6b-6c)+9(b²+c²+2bc)}
=(2a+5b+3c){(a+b)(4a-2b-6c)+9b²+9c²+18bc}
=(2a+5b+3c)(4a²+7b²+9c²+2ab+12bc-6ac)

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Question 7

$125x^3+\dfrac{1}{8}$

Sol:

⇒$5x^3+\left(\dfrac{1}{2}\right)^3$

Using identity:
x³+y³=(x+y)(x²-xy+y²)
⇒$\left(5x+\dfrac{1}{2}\right)$$+(5x)^2-(5x)\left(\dfrac{1}{2}\right)+\left(\dfrac{1}{2}\right)^2$
⇒$\left(5x+\dfrac{1}{2}\right)$$+\left(25x^2-\dfrac{5}{2}x+\dfrac{1}{4}\right)$

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Question 8

a³+b³+a+b
Sol:
⇒a³+b³+a+b
Using identity:
x³+y³=(x+y)(x²-xy+y²)
⇒(a+b)(a²-ab+b²)+(a+b)
[Taking common (a+b)]
⇒(a+b)[(a²-ab+b²)+1]
⇒(a+b)(a²-ab+b²+1)

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Question 9

x³+y³+2x²-2y²
Sol:
⇒(x³+y³)+2(x²-y²)
Using identity:
x³+y³=(x+y)(x²-xy+y²)
⇒[(x+y)(x²-xy+y²)]+2(x²-y²)
Using identity:
a²-b²=(a+b)(a-b)
⇒(x+y)(x²-xy+y²)+2(x+y)(x-y)
[Taking common (x+y)]
⇒(x+y)[(x²-xy+y²)+2(x-y)]
⇒(x+y)[x²-xy+y²+2x-2y]
⇒(x+y)(x²+y²-xy+2x-2y)

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Question 10

64x³-27y³
Sol:
⇒64x³-27y³
⇒(4x)³-(3y)³
Using identity:
x³-y³=(x-y)(x²+xy+y²)
⇒(4x-3y)[(4x)²-(4x)(3y)+(3y)²]
⇒(4x-3y)(16x²+12xy+9y²)

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Question 11

(a+b)³-(2)³
[Hint: Given expression=(a+b-2)[(a+b)²+2(a+b)+4]]
Sol:
Using identity:
x³-y³=(x-y)(x²+xy+y²)
⇒(a+b-2)[(a+b)²+2(a+b)+2²]

$=(a+b-2)\left(a^{2}+b^{2}+2 a b+2^{2}+2 a+2 b\right]$

$=(a+b-2)\left(a^{2}+b^{2}+4+2 a+2 b\right)$

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Question 12

x³y³-512
[Hint: Given expression =(xy)³-(8)³ ]
Sol :
⇒(xy)³-8³
Using identity:
x³-y³=(x-y)(x²+xy+y²)
⇒(xy-8)[(xy)²+(8)(xy)+(8)²]
⇒(xy-8)(x²y²+8xy+64)

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Question 13

$a^3-\dfrac{27}{a^3}$
Sol :
Expression$=a^3-\left(\dfrac{3}{a}\right)^3$
Using identity:
x³-y³=(x-y)(x²+xy+y²)
$=\left(a-\dfrac{3}{a}\right)\left(a^2+a.\dfrac{3}{a}+\dfrac{9}{a^2}\right)$
⇒$\left(a-\dfrac{3}{a}\right)$$+\left(a^2+a\dfrac{3}{a}+\dfrac{9}{a^2}\right)$

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Type 2
Problems based on the identity:
a³+b³+c³-3abc=(a+b+c)(a²+b²+c²-ab-bc-ca)
Category A: Problems based on factorization of thepolynomials of the form a³+b³+c³, whena+b+c=0
WORKING RULE:
Find the algebric sum of the terms. If it is zero, then required factors willbe 3×product of three terms.
Thus if a+b+c=0 , then a³+b³+c³=3abc

Category B: Problems based on factorization of thepolynomials of the forma³+b³+c³-3abc
WORKING RULE:
1. If polynomial is of the forma³+b³+c³-3abc , then use identitya³+b³+c³-3abc=(a+b+c)(a²+b²+c²-ab-bc-ca)
2. If polynomial can be converted in the forma³+b³+c³-3abc , then convert the givenexpression in this form and then factorize.
3. Use the following formulae whichever is required :
(i)a²+b²+c²-ab-bc-ca=1/2{(a-b)²+(b-c)²+(c-a)²}
(ii)(a+b+c)³=a³+b³+c³+3(a+b)(b+c)(c+a)
(iii)a³+b³+c³=(a+b+c)³-3(a+b)(b+c)(c+a)
(iv)(a+b+c)³-a³-b³-c³=3(a+b)(b+c)(c+a)

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Question 14

Fill up the blanks.

(i) If x+y+z=0 , factors of x³+y³+z³ willbe __
Sol :
Using identity :
⇒x³+y³+z³-3xyz=(x+y+z)(x²+y²+z²-xy-yz-zx)
[Given x+y+z=0]
⇒x³+y³+z³-3xyz=(0

)(x²+y²+z²-xy-yz-zx)
⇒x³+y³+z³-3xyz=0
⇒x³+y³+z³=3xyz
∴ Factors of x³+y³+z³is
⇒3xyz

(ii) Factors of (a-b)³+(b-c)³+(c-a)³ willbe __
Sol :
⇒(a-b)³+(b-c)³+(c-a)³
Using identity :
⇒a³+b³+c³=(a+b+c)³-3(a+b)(b+c)(c+a)
where a=(a-b) , b=(b-c) , c=(c-a)
⇒[(a-b)+(b-c)+(c-a)]-3[(a-b)+(b-c)][(b-c)+(c-a)][(c-a)+(a-b)]
⇒[a-b+b-c+c-a]-3[a-b+b-c][b-c+c-a][c-a+a-b]
⇒[0]-3[a-c][b-a][c-b]
⇒3(a-b)(c-a)(b-c)
⇒3(a-b)(b-c)(c-a)

(iii) Factors of (2a-3b)³+(3b-c)³+(c-2a)³will be __
Sol :

$=[2 a-3 b+3 b-c+c-2 a]^{3}-3(2 a-3 b+3 b-c)(3b-c+c-2a)(c-2a+2a-3b)$

=-3(2 a-c)(3 b-2 a)(c-3 b)

=-3×-1×-1×-1×(2a-3b)(3b-c)(c-2a)

⇒3(2a-3b)(3b-c)(c-2a)

(iv) Factors of (a+b+c)³+(a-b-c)³-8a³ willbe __
[Hint: Given expression=(a+b+c)³+(a-b-c)³+(-2a)³]
Let x=a+b+c , y=a-b-c and z=-2a
Then , x+y+z=(a+b+c)+(a-b-c)-2a=0
⇒x³+y³+z³=3xyz
⇒(a+b+c)³+(a-b-c)³-8a³
⇒3(a+b+c)(a-b-c)(-2a)
⇒-6a(a+b+c)(a-b-c)
⇒6a(a+b+c)(b+c-a)

(v) If p=2-a, prove that a³+p³-8+6ap=0

[Hint: We have, p=2-a]
⇒a+p-2=0
Let x=a , y=p and z=(-2)
Then, x+y+z=0
⇒x³+y³+z³=3xyz
⇒(a)³+(p)³+(-2)³=3×a×p×(-2)
⇒a³+p³-8=-6ap or
a³+p³+6ap-8=0

Second method:
=a³+6ap+p³-8=a³+p³+(-2)³-3ap(-2)
={a+p+(-2)}{a²+p²+(-2)²-ap-p(-2)-a(-2)}
=(a+p-2)(a²+p²+4-ap+2p+2a)
=0×(a2+p2+4-ap+2p+2a)=0

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Question 15

Factorize the following :
(i) x³-y³-z³-3xyz
Sol :

$=(x-y-z)\left(x^{2}+(-y)^{2}+(-z)^{2}-(x)(-y)-(-y)(-z)-(-z)(x)\right.$

⇒(x-y-z)(x²+y²+z²+xy+xz-yz)

(ii) a³-b³-1-3ab
Sol :

$=a^{3}-b^{3}-1^{3}-3 a b$

$[a+(-b)+(-1)]\left[a^{2}+(-b)^{2}+(-1)^{2}-(a)(-b)-(-b)(-1)-(-1)(a)\right.$
⇒(a-b-1)(a²+b²+1+ab+a-b)

(iii) a³+b³-c³+3abc
Sol :

$=(a)^{3}+(b)^{3}+(-c)^{3}-3(a)(b)-(c)$

$=(a+b-c)\left[a^{2}+b^{2}+(-c)^{2}-(a)(b)-(b)(-c)-(-c)(a)\right]$
⇒(a+b-c)(a²+b²+c²-ab+bc+ca)

(iv) x³+8y³-z³+6xyz
Sol :
$\Rightarrow(x)^{3}+2^{3} \cdot y^{3}+(-z)^{3}-3(x)(2 y)(-z)$

$\Rightarrow(x)^{3}+(2 y)^{3}+(-z)^{3}-3(x)(+2 y)(-z)$

$\Rightarrow(x+2 y-z)\left(x^{2}+(2 y)^{2}+(-z)^{2}-(x)(2 y)\right.-(2y)(-z)-(-z)(x))$

⇒(x+2y-z)(x²+4y²+z²-2xy+xz+2yz)

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Question 16

Find value of x³-8y³-36xy-216, if x=2y+6
[Hint: Given, x-2y-6=0
Let a=x , b=-2y and c=-6
Then , a+b+c=0
⇒ a³+b³+c³=3abc
⇒x³+(-2y)³+(-6)³=3x(-2y)(-6)
⇒x³-8y³-216=36xy
⇒x³-8y³-36xy-216=0

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Second method :
Given , x-2y=6
⇒(x-2y)³=(6)³
⇒x³-3x2y(x-2y)-8y³=216
⇒x³-6xy(6)-8y³-216=0
⇒x³-36xy-8y³-216=0

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Question 17

Find the value of a³+b³+c³-3abc,if
(i) a+b+c=14 and a²+b²+c²=68
[Hint: (i) a+b+c=14
⇒(a+b+c)²=196
⇒a²+b²+c²+2(ab+bc+ac)=196
⇒68+2(ab+bc+ac)=196
⇒ab+bc+ac$=\dfrac{196-68}{2}=64$
Now ,a³+b³+c³-3abc=(a+b+c)(a²+b²+c²-ab-bc-ca)
=14×(68-64)=56]
Sol :

⇒56

(ii) a+b+c=9 and ab+bc+ca=26
Sol:

$a^{3}+b^{3}+c^{3}-3 abc=(a+b+c)\left(a^{2}+b^{2}+c^{2}-a b-b c-c a\right)$

$=9\left(a^{2}+b^{2}+c^{2}-(a b+b c+c a)\right)$

$=9\left(a^{2}+b^{2}+c^{2}-26\right)$...(i)

ab+bc+ca=26

Taking square both sides

$(a+b+c)^{2}=81$

$a^{2}+b^{2}+c^{2}+2(a b+b c+ca)=81$

$a^{2}+b^{2}+c^{2}+2 \times 26=81$

$a^{2}+b^{2}+c^{2}=81-52=29$...(ii)

Putting (ii) in (i)

=9(29-26)=9×3=27

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Question 18

Find the value of x³+y³+z³ , if x+y+z=11,
x²+y²+z²=45 and xyz=40
Sol :

$x^{3}+y^{3}+z^{3}-3xy z=(x+y+z)\left(x^{2}-y^{2}+z^{2}-x y-y z-zx\right)$

$x^{3}+y^{3}+z^{3}-3 \times 40=(11)(45-(xy+yz+zx)$...(i)

x+y+z=11

$(x+y+z)^{2}=11^{2}$

$x^{2}+y^{2}+z^{2}+2\left(x y+yz+zx\right)=121$

$45+2(x y+y z+2 x)=121$

xy+yz+zx$=\frac{121-45}{2}=38$...(ii)

Putting (ii) in (i)

$x^{3}+y^{3}+z^{3}-120=11(45-38)$

$x^{3}+y^{3}+z^{3}-120=11 \times 7$

$x^{3}+y^{3}+z^{3}=77+120=197$

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Question 19

Find the product
(i) (x+y-z)(x²+y²+z²-xy+yz+zx)
Sol :

$=(x+y-z)\left(x^{2}+y^2+z^{2}-xy+y z+z x\right)$

$=[x+y+(-z)]\left[x^{2}+y^{2}+(-z)^{2}-(x) (y\right)-(y)(-z)-(-z)(x)]$

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

⇒(x³+y³-z³+3xyz)

(ii) (3x-5y-4)(9x²+25y²+15xy+12x-20y+16)
[Hint: (i) We have(x+y-z)(x²+y²+z²-xy+yz+zx)]
=[x+y+(-z)][x²+y²+(-z)²-xy-y(-z)-x(-z)]
=(a+b+c)(a²+b²+c²-ab-bc-ac) , where a=x , b=yand c=-z
=a³+b³+c³-3abc=27x³-25y³-64-3(3x)(-5y)(-4)
=27x³-125y³-64-180xy

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Question 20

(i) Find the value of x³+y³+z³ if x+y+z=1,
xy+yz+zx=-1 and xyz=-1
[Hint: (i) We know thatx³+y³+z³-3xyz
=(x+y+z)(x²+y²+z²-xy-yz-zx)]
⇒x³+y³+z³-3xyz=(x+y+z)(x²+y²+z²+2xy+2yz+2zx-3xy-3yz-3zx)
[Addingand subtracting 2xy+2yz+2zx in second bracket only]
⇒x³+y³+z³=(x+y+z)(x+y+z)2-3(xy+yz+zx)]+3xyz
[Transposing3xyz on R.H.S]
=1×[(1)²-3(-1)]+3(-1)
= 4-3 = 1

Sol :

⇒1

(ii) Simplify $\frac{\left(a^{2}-b^{2}\right)^{3}+\left(b^{2}-c^{2}\right)^{3}+\left(c^{2}-a^{2}\right)^{3}}{(a-b)^{3}+(b-c)^{3}+(c-a)^{3}}$

Sol :

(a²-b²)+(b²-c²)+(c²-a²)=0

∴Numerator
=(a²-b²)+(b²-c²)+(c²-a²)=3(a²-b²)(b²-c²)(c²-a²)
=3(a-b)(a+b)(b-c)(b+c)(c-a)(c+a)..(i)
Again, (a-b)+(b-c)+(c-a)=0
∴(a-b)³+(b-c)³+(c-a)³=3(a-b)(b-c)(c-a)..(ii)
Hence, given expression=(a+b)(b+c)(c+a)

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Question 21

Prove that:
(i) (a³+b³+c³)-3abc=1/2(a+b+c){(a-b)²+(b-c)²+(c-a)²}
Sol :

$a^{3}+b^{3}+c^{3}-3 a b c=(a+b+c)\left(a^{2}+b^{2}+c^{2}-ab-bc-ca\right)$

$=\frac{1}{2} \times(a+b+c) \left\{a^{2}+b^{2}-2 a b+b^{2}+c^{2}-2 bc\right.\left.+c^{2}+a^{2}-2 c a\right\}$

$=\frac{1}{2} \times (a+b+c) \left(2 a^{2}+2 b^{2}+2 c^{2}-2 a b\right. -2 bc-2 c a)$

$=2\times \frac{1}{2}(a+b+c)\left\{\left(a^{2}+b^{2}+c^{2}-a b-b c-ca\right\}\right.$

$a^{2}+b^{3}+c^{3}-3 a b c$

(ii) (a+b)³+(b+c)³+(c+a)³-3(a+b)(b+c)(c+a)=2(a³+b³+c³-3abc)

Sol :

$\Rightarrow(a+b)^{3}+(b+c)^{3}+(c+a)^{3}-3(a+b)(b+c)(c+a)$

$\left[(a+b)+(b+c)+(c+a)]\left((a+b)^{2}+(b+c)^{2}+\left(c+a)^{2}\right.\right.\right.-(a+b)(b+c)-(b+c)(c+a)-((c+a)(a+b)\}$

$\Rightarrow(2 a+2 b+2 c)\left\{a^{2}+b^{2}+2 a b+b^{2}+c^{2}+2 bc+c+a^{2}\right.+2 c a-[a(b+c)+b(b+c)]-[b(c+a)+c(c+a)]-[c(a+b)+a(a+b)]$

$=2(a+b+c) \{2 a^{2}+2 b^{2}+2 c^{2}+2 a b+2 b c+2 c a-\left[a b+a c+b^{2}+bc\right]-\left[b c+b a+c^{2}+c a\right]-\left[c a+c b+a^{2}\right.+ab]$

$\Rightarrow 2(a+b+c)\left[2 a^{2}+2 b^{2}+2 c^{2}+2 a b+2 b c+2 c a-a b-ac-b^{2}-b c\right. \left.-b c-b a-c^{2}-c a-c a-c b-a^{2}-a b\right\}$

$\Rightarrow 2(a+b+c)\left\{a^{2}+b^{2}+c^{2}+a b+b c+c a-a c-b c-b a-ca-cb-ab\right\}$

$\Rightarrow \quad 2(a+b+c)\left(a^{2}+b^{2}+c^{2}-c a-c b-a b\right)$

=2(a³+b³+c³-3abc)

(iii) Factorize :p³(q-r)³+q³(r-p)³+r³(p-q)³

Sol :

$\Rightarrow[p(q-r)]^{3}+[q(r-p)]^{3}+[r(p-q)]^{3}$

[p(q-r)]=x , [q(r-p)]=y , [r(p-q)]=z

x³+y³+z³=3xyz, when x+y+z=0

=[p(q-r)]+[q(r-p)]+[r(p-q)]

=pq-rp+rq-qp+rp-rq

∴3[p(q-r)]+[q(r-p)]+[r(p-q)]

(iv) Find the value of(x-a)³+(x-b)³+(x-c)³=3(x-a)(x-b)(x-c) whena+b+c=3x
Sol :

$(x-a)^{3}+(x-b)^{3}+(x-c)^{3}=3(x-a)(x-b)(x-c)$

$(x-a)^{3}+(x-b)^{3}+(x-c)^{3}-3(x-a)(x-b)(x-c)$

Using identity

$x^{3}+y^{3}+z^{3}-3 x y z=(x+y+z)\left(x^{2}+y^{2}+z^{2}-x y-y z-z x\right)$

$=[(x-a)+(x-b)+(x-c)]\left[(x-a)^{2}+(x-b)^{2}+(x-c)^{2}-(x-a)(x-b)-(x-b)(x-c)-(x-c)(x-a)\right]$

$=[x-a+x-b+x-c]\left[x^{2}+a^{2}-2 a x+x^{2}+b^{2}-2 x b+x^{2}+c^{2}\right.-2x c-[x(x-b)-a(x-b)]-[x(x-c)-b(x-c)]-[x(x-a)-c(x-a)]$

$\Rightarrow[3x-a-b-c]\left\{3x^{2}+a^{2}-2xa+b^{2}-2xb+c^{2}-2xc-[x^2-xb-xa+ab]-[x^2-xc-bx+bc]-[x^2-xa-cx+ca]\right\}$

$\Rightarrow[3x-(a+b+c)]\left\{3x^{2}+a^{2}-2xa+b^{2}-2xb+c^{2}-2xc-[x^2-xb-xa+ab]-[x^2-xc-bx+bc]-[x^2-xa-cx+ca]\right\}$

∵a+b+c=3x

$\Rightarrow[3x-(3x)]\left\{3x^{2}+a^{2}-2xa+b^{2}-2xb+c^{2}-2xc-[x^2-xb-xa+ab]-[x^2-xc-bx+bc]-[x^2-xa-cx+ca]\right\}$

$\Rightarrow[0]\times\left\{3x^{2}+a^{2}-2xa+b^{2}-2xb+c^{2}-2xc-[x^2-xb-xa+ab]-[x^2-xc-bx+bc]-[x^2-xa-cx+ca]\right\}$