Q.1. Factorize: x^{3} + 27
Ans. x^{3}+27=(x)^{3}+(3)^{3}
=(x+3)(x^{2}−3x+3^{2})
=(x+3)(x^{2}−3x+9)
Q.2. Factorise: 27a^{3} + 64b^{3}
Ans. We know that x^{3}+y^{3}
=(x+y)(x^{2}+y^{2}−xy)x^{3}+y^{3}
Given: 27a^{3} + 64b^{3}
x = 3a, y = 4b
27a^{3} + 64b^{3}=(3a+4b)(9a^{2}+16b^{2}−12ab)
Q.3. Factorize: 125a^{3}+1/8
Ans. 125a3+1/8=(5a)^{3}+(1/2)^{3}
=(5a+1/2)[(5a)^{2}−5a×1/2+(1/2)^{2}]
=(5a+1/2)(25a^{2}−5a/2+1/4)
Q.4. Factorize: 216x^{3}+1/125
Ans. 216x^{3}+1/125=(6x)^{3}+(1/5)^{3}
=(6x+1/5)[(6x)^{2}−6x×1/5+(1/5)^{2}]
=(6x+1/5)(36x^{2}−6x/5+1/25)
Q.5. Factorize: 16x^{4} + 54x
Ans. 16x^{4}+54x
=2x(8x^{3}+27)
=2x[(2x)^{3}+(3)^{3}]
=2x(2x+3)[(2x)^{2}−2x×3+3^{2}]
=2x(2x+3)(4x^{2}−6x+9)
Q.6. Factorize: 7a^{3} + 56b^{3}
Ans. 7a^{3}+56b^{3}=7(a^{3}+8b^{3})
=7[(a)^{3}+(2b)^{3}]
=7(a+2b)[a^{2}−a×2b+(2b)^{2}]
=7(a+2b)(a^{2}−2ab+4b^{2})
Q.7. Factorize: x^{5} + x^{2}
Ans. x^{5}+x^{2}=x^{2}(x^{3}+1)
=x^{2}(x^{3}+1^{3})
=x^{2}(x+1)(x^{2}−x×1+1^{2})
=x^{2}(x+1)(x^{2}−x+1)
Q.8. Factorize: a^{3} + 0.008
Ans. a^{3}+0.008=a^{3}+(0.2)^{3}
=(a+0.2)[a^{2}−a×(0.2)+(0.2)^{2}]
=(a+0.2)(a^{2}−0.2a+0.04)
Q.9. Factorise: 1 – 27a^{3}
Ans. 1−27a^{3}=1^{3}−(3a)^{3}
=(1−3a)[1^{2}+1×3x+(3a)^{2}]
=(1−3a)(1+3a+9a^{2})
Q.10. Factorize: 64a^{3 }− 343
Ans. 64a^{3}−343=(4a)^{3}−(7)^{3}
=(4a−7)(16a^{2}+4a×7+7^{2})
=(4a−7)(16a^{2}+28a+49)
Q.11. Factorize: x^{3} − 512
Ans. x^{3}−512 =x^{3}−8^{3}
=(x−8)(x^{2}+8x+8^{2})
=(x−8)(x^{2}+8x+64)
Q.12. Factorize: a^{3} − 0.064
Ans. a^{3}−0.064=(a)^{3}−(0.4)^{3}
=(a−0.4)[a^{2}+a×(0.4)+(0.4)^{2}]
=(a−0.4)(a^{2}+0.4a+0.16)
Q.13. Factorize: 8x^{3}−1/27y^{3}
Ans. 8x^{3}−127y^{3}=(2x)^{3}−(1/3y)^{3}
=(2x−1/3y)[(2x)^{2}+2x×1/3y+(1/3y)^{2}]
=(2x−1/3y)(4x^{2}+2x/3y+1/9y^{2})
Q.14. Factorise: x^{3}/216−8y^{3}
Ans. We know a^{3}−b^{3}=(a−b)(a^{2}+b^{2}+ab)
We have,
x^{3}/216−8y^{3}=(x/6)^{3}−(2y)^{3}
So, a=x/6,b=2y
x^{3}/216−8y^{3}=(x/6−2y)((x/6)^{2}+x/6×2y+(2y)^{2})
=(x/6−2y)(x^{2}/36+xy/3+4y^{2})
Q.15. Factorize: x − 8xy^{3}
Ans. x−8xy^{3}=x(1−8y^{3})
=x[1^{3}−(2y)^{3}]
=x(1−2y)(1^{2}+1×2y+(2y)^{2})
=x(1−2y)(1+2y+4y^{2})
Q.16. Factorise: 32x^{4} – 500x
Ans. 32x^{4} – 500x=4x(8x^{3}−125)=4x((2x)^{3}−5^{3})
we know
a^{3}−b^{3}=(a−b)(a^{2}+b^{2}+ab)
a=2x,b=5
32x^{4} – 500x=4x((2x)^{3}−5^{3})
=4x(2x−5)(4x^{2}+25+10x)
Q.17. Factorize: 3a^{7}b − 81a^{4}b^{4}
Ans. 3a^{7}b−81a^{4}b^{4}=3a^{4}b(a^{3}−27b^{3})
=3a^{4}b[a^{3}−(3b)^{3}]
=3a^{4}b(a−3b)[a^{2}+a×3b+(3b)^{2}]
=3a^{4}b(a−3b)(a^{2}+3ab+9b^{2})
Q.18. Factorise: x4 y4 – xy
Ans. Using the identity
a^{3}−b^{3}=(a−b)(a^{2}+b^{2}+ab)
x^{4} y^{4}–xy=xy(x^{3}y^{3}−1)
=xy(xy−1)(x^{2}y^{2}+1+xy)
Q.19. Factorise: 8x^{2} y^{3 }– x^{5}
Ans. 8x^{2}y^{3}–x^{5}=x^{2}(8y^{3}−x^{3})
=x^{2}(2y−x)(4y^{2}+x2+2xy)
Q.20. Factorise:1029 – 3x^{3}
Ans. 1029–3x^{3}
=3(343−x^{3})
=3(7^{3}−x^{3})
=3(7−x)(49+x^{2}+7x)
Q.21. Factorize: x^{6} − 729
Ans. x^{6}−729=(x^{2})^{3}−(9)^{3}
=[x^{2}−9][(x^{2})^{2}+x^{2}×9+9^{2}]
=[x^{2}−3^{2}](x^{4}+9x^{2}+81)
=(x+3)(x−3)(x^{4}+18x^{2}+81−9x^{2})
=(x+3)(x−3)[(x^{2})^{2}+2×x^{2}×9+9^{2}−9x^{2}]
=(x+3)(x−3)[(x^{2}+9)^{2}−(3x)^{2}]
=(x+3)(x−3)(x^{2}+9+3x)(x^{2}+9−3x)
=(x+3)(x−3)(x^{2}+3x+9)(x^{2}−3x+9)
Q.22. Factorise: x^{9} – y^{9}
Ans. x^{9}–y^{9}=(x^{3})^{3}−(y^{3})^{3}
we know
a^{3}−b^{3}=(a−b)(a^{2}+b^{2}+ab)
a=x^{3},b=y^{3}
So,x^{9}–y^{9}=(x^{3})^{3}−(y^{3})^{3}
=(x^{3}−y^{3})(x^{6}+y^{6}+x^{3}y^{3})
=(x−y)(x^{2}+y^{2}+xy)(x^{6}+y^{6}+x^{3}y^{3})
Q.23. Factorize: (a + b)^{3} − (a − b)^{3}
Ans. (a + b)^{3}−(a−b)^{3}=[(a+b)−(a−b)][(a+b)^{2}+(a+b)(a−b)+(a−b)^{2}]
=(a+b−a+b)[a^{2}+2ab+b^{2}+a^{2}−b^{2}+a^{2}−2ab+b^{2}]
=2b(3a^{2}+b^{2})
Q.24. Factorize: 8a^{3} − b^{3} − 4ax + 2bx
Ans. 8a^{3}−b^{3}−4ax+2bx=[(2a)^{3}−(b)^{3}]−2x(2a−b)
=(2a−b)[(2a)^{2}+2ab+b^{2}]−2x(2a−b)
=(2a−b)(4a^{2}+2ab+b^{2})−2x(2a−b)
=(2a−b)(4a^{2}+2ab+b^{2}−2x)
Q.25. Factorize: a^{3} + 3a^{2}b + 3ab^{2} + b^{3} − 8
Ans. a^{3}+3a^{2}b+3ab^{2}+b^{3}−8=(a^{3}+b^{3}+3a^{2}b+3ab^{2})−8
=[a^{3}+b^{3}+3ab(a+b)]−8
=(a+b)^{3}−2^{3}
=(a+b−2)[(a+b)^{2}+2(a+b)+2^{2}]
=(a+b−2)[(a+b)^{2}+2(a+b)+4]
Q.26. Factorize: a^{3}−1/a^{3}−2a+2/a
Ans. a^{3}−1/a^{3}−2a+2/a=(a^{3}−1/a^{3})−2(a−1/a)
=[(a)^{3}−(1/a)^{3}]−2(a−1/a)
=(a−1/a)[a^{2}+a×1/a+(1/a)2]−2(a−1/a)
=(a−1/a)(a^{2}+1+1/a^{2})−2(a−1/a)
=(a−1/a)(a^{2}+1+1/a^{2}−2)
=(a−1/a)(a^{2}−1+1/a^{2})
Q.27. Factorize: 2a^{3} + 16b^{3} − 5a − 10b
Ans. 2a^{3}+16b^{3}−5a−10b=2[a^{3}+8b^{3}]−5(a+2b)
=2[a^{3}+(2b)^{3}]−5(a+2b)
=2(a+2b)[a^{2}−a×2b+(2b)^{2}]−5(a+2b)
=2(a+2b)(a^{2}−2ab+4b^{2})−5(a+2b)
=(a+2b)[2(a^{2}−2ab+4b^{2})−5]
Q.28. Factorise: a^{6} + b^{6}
Ans. a^{6}+b^{6}=(a^{2})^{3}+(b^{2})^{3}
=(a^{2}+b^{2})[(a^{2})^{2}−a^{2}b^{2}+(b^{2})^{2}]
=(a^{2}+b^{2})(a^{4}−a^{2}b^{2}+b^{4})
Q.29. Factorise: a^{12} – b^{12}
Ans. a^{12} – b^{12}
=(a^{6}+b^{6})(a^{6}−b^{6})
=[(a^{2})^{3}+(b^{2})^{3}][(a^{3})^{2}−(b^{3})^{2}]
=[(a^{2}+b^{2})(a^{4}+b^{4}−a^{2}b^{2})][(a^{3}−b^{3})(a^{3}+b^{3})]
=[(a^{2}+b^{2})(a^{4}+b^{4}−a^{2}b^{2})][(a−b)(a^{2}+b^{2}+ab)(a+b)(a^{2}+b^{2}−ab)]
=(a−b)(a^{2}+b^{2}+ab)(a+b)(a^{2}+b^{2}−ab)(a^{2}+b^{2})(a^{4}+b^{4}−a^{2}b^{2})
Q.30. Factorise: x^{6} – 7x^{3} – 8
Ans. Let x^{3}=y
So, the equation becomes
y^{2}−7y−8=y^{2}−8y+y−8
=y(y−8)+(y−8)
=(y−8)(y+1)
=(x^{3}−8)(x^{3}+1)
=(x−2)(x^{2}+4+2x)(x+1)(x^{2}+1−x)
Q.31. Factorise: x^{3} – 3x2 + 3x + 7
Ans. x^{3} – 3x^{2} + 3x + 7
=x3–3x^{2}+3x+7
=x^{3}–3x^{2}+3x+8−1
=x^{3}–3x^{2}+3x−1+8
=(x^{3}–3x^{2}+3x−1)+8
=(x−1)^{3}+2^{3}
=(x−1+2)[(x−1)^{2}+4−2(x−1)]
=(x+1)[x^{2}+1−2x+4−2x+2]
=(x+1)(x^{2}−4x+7)
Q.32. Factorise: (x +1)^{3} + (x – 1)^{3}
Ans. (x +1)^{3} + (x – 1)^{3}
=(x+1+x−1)[(x+1)2+(x−1)2−(x−1)(x+1)]
=(2x)[(x+1)^{2}+(x−1)^{2}−(x^{2}−1)]
=2x(x^{2}+1+2x+x^{2}+1−2x−x^{2}+1)
=2x(x2+3)
Q.33. Factorise: (2a +1)^{3} + (a – 1)^{3}
Ans. (2a +1)^{3} + (a – 1)^{3}
=(2a+1+a−1)[(2a+1)^{2}+(a−1)^{2}−(2a+1)(a−1)]
=(3a)[4a^{2}+1+4a+a^{2}+1−2a−2a^{2}+2a−a+1]
=3a[3a^{2}+3a+3]
=9a(a^{2}+a+1)
Q.34. Factorise: 8(x +y)^{3} – 27(x – y)^{3}
Ans. 8(x +y)^{3} – 27(x – y)^{3}
=[2(x+y)]^{3}−[3(x−y)]^{3}
=(2x+2y−3x+3y)[4(x+y)^{2}+9(x−y)^{2}+6(x^{2}−y^{2})]
=(−x+5y)[4(x^{2}+y^{2}+2xy)+9(x^{2}+y^{2}−2xy)+6(x^{2}−y^{2})]
=(−x+5y)[4x^{2}+4y^{2}+8xy+9x^{2}+9y^{2}−18xy+6x^{2}−6y^{2}]
=(−x+5y)(19x^{2}+7y^{2}−10xy)
Q.35. Factorise: (x +2)^{3} + (x – 2)^{3}
Ans. (x +2)^{3} + (x – 2)^{3}
=(x+2+x−2)[(x+2)^{2}+(x−2)^{2}−(x^{2}−4)]
=2x(x^{2}+4+4x+x^{2}+4−4x−x2+4)
=2x(x^{2}+12)
Q.36. Factorise: (x + 2)^{3} – (x – 2)^{3}
Ans. (x + 2)^{3} – (x – 2)^{3}
=(x+2−x+2)[(x+2)^{2}+(x−2)^{2}+(x^{2}−4)]
=4[x^{2}+4+4x+x^{2}+4−4x+x^{2}−4]
=4(3x^{2}+4)
Q.37. Prove that (0.85×0.85×0.85+0.15×0.15×0.15)/(0.85×0.85−0.85×0.15+0.15×0.15)=1.
Ans. LHS: (0.85×0.85×0.85+0.15×0.15×0.15)/(0.85×0.85−0.85×0.15+0.15×0.15)
=(0.85)^{3}+(0.15)^{3}/(0.85)^{2}−0.85×0.15+(0.15)^{2}
We know a^{3}+b^{3}=(a+b)(a^{2}+b^{2}−ab)
Here a=0.85,b=0.15
(0.85)^{3}+(0.15)^{3}/(0.85)^{2}−0.85×0.15+(0.15)^{2}
=(0.85+0.15)((0.85)2−0.85×0.15+(0.15)2)(0.85)2−0.85×0.15+(0.15)2=0.85+0.15=1:RHS
Thus, LHS = RHS
Q.38. Prove that (59×59×59−9×9×9)/(59×59+59×9+9×9)=50.
Ans. (59×59×59−9×9×9)/(59×59+59×9+9×9)
=(59)^{3}−9^{3}/59^{2}+59×9+9^{2}
We know a^{3}+b^{3}=(a+b)(a^{2}+b^{2}−ab)
Here a=59,b=9
So, ((59−9)(59^{2}+9^{2}+59×9))/(59^{2}+9^{2}+59×9)=59−9=50:RHS
Thus, LHS=RHS
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