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Class 9 Exam  >  Class 9 Notes  >  Extra Documents & Tests for Class 9  >  RS Aggarwal Solutions: Exercise 3G - Factorisation of Polynomials

RS Aggarwal Solutions: Exercise 3G - Factorisation of Polynomials | Extra Documents & Tests for Class 9 PDF Download

Q.1. Find the product: (x + y − z) (x2 + y2 + z2 − xy + yz + zx)
Ans. (x+y−z)(x2+y2+z2−xy+yz+zx)
=[x+y+(−z)][x2+y2+(−z)2−xy−y×(−z)−[−z]×x]
=x3+y3+(−z)3−3x×y×(−z)
=x3+y3−z3+3xyz

Q.2. Find the product: (x – y − z) (x2 + y2 + z2 + xy – yz + xz)
Ans. (x – y − z) (x2 + y2 + z2 + xy – yz + xz)
=(x+(−y)+(−z)) (x2+y2+z2+xy–yz+xz)
We know
(a+b+c)(a2+b2+c2−ab−bc−ca)
=a3+b3+c3−3abc
Here, a=x,b=−y,c=−z
(x+(−y)+(−z)) (x2+y2+z2+xy–yz+xz)
=x3−y3−z3−3xyz

Q.3. Find the product: (x − 2y + 3) (x2 + 4y2 + 2xy − 3x + 6y + 9)
Ans. (x − 2y + 3)(x2+4y2+2xy−3x+6y+9)
=(x − 2y + 3)(x2+4y2+9+2xy+6y−3x)
=[x+(−2y)+3][x2+(−2y)2+(3)2−x×(−2y)−(−2y)×3−3×x]
=(x)3+(−2y)3+(3)3−3(x)(−2y)(3)
=x3−8y3+27+18xy

Q.4. Find the product: (3x – 5y + 4) (9x2 + 25y2 + 15xy − 20y + 12x + 16)
Ans. (3x−5y+4)(9x2+25y2+15xy−20y+12x+16)
=(3x+(−5y)+4)(9x2+25y2+16+15xy−20y+12x)
(a+b+c)(a2+b2+c2−ab−bc−ca)
=a3+b3+c3−3abc
Here, a=3x,b=−5y,c=4
(3x+(−5y)+4)(9x2+25y2+16+15xy−20y+12x)
=(3x)3+(−5y)3+43−3×3x(−5y)(4)
=27x3−125y3+64+180xy

Q.5. Factorize: 125a3 + b3 + 64c3 − 60abc
Ans. 125a3+b3+64c3−60abc=(5a)3+(b)3+(4c)3−3×5a×b×4c
=(5a+b+4c)[(5a)2+(b)2+(4c)2−5a×b−b×4c−5a×4c]
=(5a+b+4c)(25a2+b2+16c2−5ab−4bc−20ac)

Q.6. Factorize: a3 + 8b3 + 64c3 − 24abc
Ans. a3+8b3+64c3−24abc=a3+(2b)3+(4c)3−3×a×2b×4c
=(a+2b+4c)[a2+(2b)2+(4c)2−a×2b−2b×4c−4c×a]
=(a+2b+4c) (a2+4b2+16c2-2ab-8bc-4ca)

Q.7. Factorize: 1 + b3 + 8c3 − 6bc
Ans. 1+b3+8c3−6bc=(1)3+(b)3+(2c)3−3×1×b×2c
=(1+b+2c)[12+b2+(2c)2−1×b−b×2c−1×2c]
=(1+b+2c)(1+b2+4c2−b−2bc−2c)

Q.8. Factorize: 216 + 27b3 + 8c3 − 108abc
Ans. 216+27b3+8c3−108abc
=(6)3+(3b)3+(2c)3−3×6×3b×2c
=(6+3b+2c)[62+(3b)2+(2c)2−6×3b−3b×2c−2c×6]
=(6+3b+2c)(36+9b2+4c2−18b−6bc−12c)

Q.9. Factorize: 27a3 − b3 + 8c3 + 18abc
Ans. 27a3−b3+8c3+18abc=(3a)3+(−b)3+(2c)3−3×(3a)×(−b)×(2c)
=[3a+(−b)+2c][(3a)2+(−b)2+(2c)2−3a(−b)−(−b)2c−3a×2c]
=(3a−b+2c)(9a2+b2+4c2+3ab+2bc−6ac)

Q.10. Factorize: 8a3 + 125b3 − 64c3 + 120abc
Ans. 8a3+125b3−64c3+120abc=(2a)3+(5b)3+(−4c)3−3×(2a)×(5b)×(−4c)
=(2a+5b−4c)[(2a)2+(5b)2+(−4c)2−(2a)(5b)−(5b)(−4c)−(2a)×(−4c)]
=(2a+5b−4c)(4a2+25b2+16c2−10ab+20bc+8ac)

Q.11. Factorize: 8 − 27b3 − 343c3 − 126bc
Ans. 8−27b3−343c3−126bc
=(2)3+(−3b)3+(−7c)3−3×(2)×(−3b)×(−7c)
=[2+(−3b)+(−7c)][(2)2+(−3b)2+(−7c)2−(2)(−3b)−(−3b)(−7c)−(2)(−7c)]
=(2−3b−7c)(4+9b2+49c2+6b−21bc+14c)

Q.12. Factorize: 125 − 8x3 − 27y3 − 90xy
Ans. 125−8x3−27y3−90xy=53+(−2x)3+(−3y)3−3×5×(−2x)×(−3y)
=[5+(−2x) +(−3y)][52+(−2x)2+(−3y)2−5×(−2x)−(−2x)(−3y)−5×(−3y)]                                     =(5−2x−3y)(25+4x2+9y2+10x−6xy+15y)

Q.13. Factorize: 2√2a3 + 16√2b3+c3−12abc
Ans. 2√2a3+16√2b3+c3−12abc
=(√2a)3+(2√2b)3+c3−3×(√2a)×(2√2b)×(c)
=(√2a+2√2b+c)[(√2a)2+(2√2b)2+c2−(√2a)×(2√2b)−(2√2b)×(c)−(2√a)×(c)]
=(√2a+2√2b+c)(2a2+8b2+c2−4ab−2√2bc−√2ac)

Q.14. Factorise: 27x3 – y3 – z3 – 9xyz
Ans. 27x3−y3–z3–9xyz
=(3x)3−y3−z3−3×(3x)×(−y)×(−z)
We know,
a3+b3+c3−3abc
=(a+b+c)(a2+b2+c2−ab−bc−ca)
a=3x,b=−y,c=−z
(3x)3−y3−z3−3×(3x)×(−y)×(−z)
=(3x−y−z)(9x2+y2+z2+3xy−yz+3xz)

Q.15. Factorise: 2√2a3+3√3b3+c3−3√6abc
Ans. 2√2a3+3√3b3+c3−3√6abc
=(√2a)3+(√3b)3+c3−3(√2a)(√3b)c
We know
x3+y3+z3−3xyz=(x+y+z)(x2+y2+z2−xy−yz−zx)
x=√2a,y=√3b,z=c
(√2a)3+(√3b)3+c3−3(√2a)(√3b)c
=(√2a+√3b+c)(2a2+3b2+c2−RS Aggarwal Solutions: Exercise 3G - Factorisation of Polynomials | Extra Documents & Tests for Class 9 - √2ac)

Q.16. Factorise: 3√3a3−b3−5√5c3−3√15abc
Ans. 3√3a3−b3−5√5c3−3√15abc
=(√3a)3+(−b)3+(−√5c)3−3(√3a)(−b)(−√5c)
We know
x3+y3+z3−3xyz=(x+y+z)(x2+y2+z2−xy−yz−zx)
Here, x=(√3a),y=(−b),z=(−√5c)
3√3a3−b3−5√5c3−3√15abc
=(√3a)3+(−b)3+(−√5c)3−3(√3a)(−b)(−√5c)
=(√3a−b−√5c)(3a2+b2+5c2+√3ab−√5bc+√15c)

Q.17. Factorize: (a − b)3 + (b − c)3 + (c − a)3
Ans. (a−b)3+(b−c)3+(c−a)3
Putting (a−b)=x, (b−c)=y and (c−a)=z, we get:
(a−b)3+(b−c)3+(c−a)3
=x3+y3+z3[Where (x+y+z)=(a−b)+(b−c)+(c−a)=0]
=3xyz [(x+y+z)=0 ⇒x3+y3+z3=3xyz]
=3(a−b)(b−c)(c−a)

Q.18. Factorise: (a−3b)3+(3b−c)3+(c−a)3
Ans. We know x3+y3+z3−3xyz=(x+y+z)(x2+y2+z2−xy−yz−zx)
x3+y3+z3=(x+y+z)(x2+y2+z2−xy−yz−zx)+3xyz
Here, x=(a−3b),y=(3b−c),z=(c−a)
(a−3b)3+(3b−c)3+(c−a)3
=(a−3b+3b−c+c−a)[(a−3b)2+(3b−c)2+(c−a)2−(a−3b)(3b−c)−(3b−c)(c−a)−(c−a)(a−3b)]+3(a−3b)(3b−c)(c−a)
=0+3(a−3b)(3b−c)(c−a)
=3(a−3b)(3b−c)(c−a)

Q.19. Factorize: (3a − 2b)3 + (2b − 5c)3 + (5c − 3a)3
Ans. Put (3a−2b)=x, (2b−5c)=y and (5c−3a)=z.
We have:
x+y+z = 3a−2b+2b−5c+5c−3a=0
Now,(3a−2b)3+(2b−5c)3+(5c−3a)3=x3+y3+z3
=3xyz [Here, x+y+z=0. So, x3 + y3 +z3 = 3xyz]
=3(3a−2b)(2b−5c)(5c−3a)

Q.20. Factorize: (5a − 7b)3 + (9c − 5a)3 + (7b − 9c)3
Ans. Put (5a−7b)=x, (9c−5a)=z and (7b−9c)=y.
Here, x+y+z = 5a − 7b + 9c−5a+7b−9c=0
Thus, we have:
(5a−7b)3+(9c−5a)3+(7b−9c)3=x3+z3+y3
=3xzy   [When x+y+z=0, x3+y3+z3 = 3xyz.]
=3 (5a−7b)(9c−5a)(7b−9c)

Q.21. Factorize: a3(b − c)3 + b3(c − a)3 + c3(a − b)3
Ans. We have: a3(b−c)3+b3(c−a)3+c3(a−b)3 = [a(b−c)]3+[b(c−a)]3+[c(a−b)]3
Put a(b−c) = x
b(c−a) = y
c(a−b) = z
Here, x+y+z = a(b−c)+b(c−a)+c(a−b)
=ab − ac + bc − ab + ac − bc
=0
Thus, we have:
a3(b−c)3+b3(c−a)3+c3(a−b)3 =x3 + y3+ z3
=3xyz [When x+y+z =0, x3 + y3+ z3 =3xyz.]
=3 a(b−c)b(c−a)c(a−b)
=3abc(a−b)(b−c)(c−a)

Q.22. Evaluate
(i) (–12)3 + 73 + 53
(ii) (28)3 + (–15)3 + (–13)3
Ans. (i) (–12)3 + 73 + 53
We know
x3+y3+z3−3xyz=(x+y+z)(x2+y2+z2−xy−yz−zx)
x3+y3+z3=(x+y+z)(x2+y2+z2−xy−yz−zx)+3xyz
Here, x=(−12),y=7,z=5
(−12)3+73+53
=(−12+7+5)[(−12)2+72+52−7(−12)−35+60]+3(−12)×35
=0−1260
= -1260
(ii) (28)3 + (–15)3 + (–13)3
We know
x3+y3+z3−3xyz=(x+y+z)(x2+y2+z2−xy−yz−zx)
x3+y3+z3=(x+y+z)(x2+y2+z2−xy−yz−zx)+3xyz
Here, x=(−28),y=−15,z=−13
(28)3+(−15)3+(−13)3
=(28−15−13)[(28)2+(−15)2+(−13)2−28(−15)−(−15)(−13)−28(−13)]+3×28(−15)(−13)
=0+16380=16380

Q.23. Prove that (a+b+c)3−a3−b3−c3=3(a+b) (b+c) (c+a)
Ans. (a+b+c)3=[(a+b)+c]3
= (a+b)3+c3+3(a+b)c(a+b+c)
⇒(a+b+c)3=a3+b3+3ab(a+b)+c3+3(a+b)c(a+b+c)
⇒(a+b+c)3−a3+b3−c3=3ab(a+b)+3(a+b)c(a+b+c)
⇒(a+b+c)3−a3+b3−c3=3(a+b)[ab+ca+cb+c2]
⇒(a+b+c)3−a3+b3−c3=3(a+b)[a(b+c)+c(b+c)]
⇒(a+b+c)3−a3+b3−c3=3(a+b)(b+c)(a+c)

Q.24. If a, b, c are all nonzero and a + b + c = 0, prove that a2/bc+b2/ca+c2/ab=3.
Ans. a+b+c=0
⇒a3+b3+c3=3abc
Thus, we have:
(a2/bc+b2/ca+c2/ab)=(a3+b3+c3)/abc
=3abc/abc
=3

Q.25. If a + b + c = 9 and a2 + b2 + c2 = 35, find the value of (a3 + b3 + c3 – 3abc).
Ans. a + b + c = 9
⇒(a+b+c)2=92=81
⇒a2+b2+c2+2(ab+bc+ca)=81
⇒35+2(ab+bc+ca)=81
⇒(ab+bc+ca)=23
We know,
(a3 + b3 + c3 – 3abc)
= (a+b+c)(a2+b2+c2−ab−bc−ca)
=(9)(35−23)
=108

The document RS Aggarwal Solutions: Exercise 3G - Factorisation of Polynomials | Extra Documents & Tests for Class 9 is a part of the Class 9 Course Extra Documents & Tests for Class 9.
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