Expand: (i) (2x-y^2)^3 (ii) (2x+1/2x)^3 (iii) (x^2-y^2)^3 (iv) (4a2-3b)^3 or Find Cube of Each of The Following. [S.Chand ICSE Class 9 Fundamentals of Mathematics Chapter 3 Exercise 3.1 Expansion and Factorisation Solution]
(i) (2x-y2)3
= (2x)3 -3.(2x)2.(y2) +3.(3x).(y2)2 -(y2)3
= 8x3 -3. 4x2 .y2 +3. 3x . y4 -y6
=8x3 -12x2y2 +8xy4 -y6 [Ans]
(ii) $\left(2{\mathrm{x}} + \frac{1}{2{\mathrm{x}}}\right)^3$
= $\left(2{\mathrm{x}}\right)^3 + 3.\left(2{\mathrm{x}}\right)^2\ldotp \frac{1}{2{\mathrm{x}}} + 3.(2{\mathrm{x}})\ldotp \frac{1}{\left(2{\mathrm{x}}\right)^2} + \left(\frac{1}{2{\mathrm{x}}}\right)^3$
= $8{\mathrm{x}}^3 + 3.4{\mathrm{x}}^2\ldotp \frac{1}{2{\mathrm{x}}} + 3.2{\mathrm{x}}\ldotp \frac{1}{4{\mathrm{x}}^2} + \frac{1}{8{\mathrm{x}}^3}$
= $8{\mathrm{x}}^3 + 6{\mathrm{x}} + \frac{3}{2{\mathrm{x}}} + \frac{1}{8{\mathrm{x}}^3}$ [Ans]
(iii) Expand: (x2-y2)3
= (x2)3 -3(x2)2 (y2)+3.(x2)(y2)2-(y2)3
=x6-3x4y2+3x2y4-y6 [Ans]
(iv) (4a2-3b)3
=(4a2)3 -3.(4a2)2(3b)+3.(4a2)(3b)2 -(3b)3
= 64a6-3.16a4 .3b + 3. 4a2 .9b2 -27b3
=64a6 – 144a4b+ 108a2b2 -27b3 [Ans]