Write the following cubes in expanded form:
(i) (2x + 1)3
(ii) (2a − 3b)3
(iii) ((3/2)x + 1)3
(iv) (x−(2/3)y)3
(i) (2x+1)3
Using identity,(x+y)3 = x3+y3+ 3xy(x+y)
(2x+1)3= (2x)3+13+(3×2x×1)(2x+1)
= 8x3+1+6x(2x+1)
= 8x3+12x2+6x+1
(ii) (2a−3b)3
Using identity,(x–y)3 = x3–y3–3xy(x–y)
(2a−3b)3 = (2a)3−(3b)3–(3×2a×3b)(2a–3b)
= 8a3–27b3–18ab(2a–3b)
= 8a3–27b3–36a2b+54ab2
(iii) ((3/2)x+1)3
Using identity,(x+y)3 = x3+y3+3xy(x+y)
((3/2)x+1)3=((3/2)x)3+13+(3×(3/2)x×1)((3/2)x +1)
= \(\frac{27}{8}x^3 + 1 + \) \(\frac{9}{2}x(\frac{3}{2}x + 1)\)
= \(\frac{27}{8}x^3 + 1 + \) \(\frac{27}{4}x^2 + \frac{9}{2}x \)
= \(\frac{27}{8}x^3 + 1 + \) \(\frac{27}{4}x^2 + \frac{9}{2}x + 1\)
(iv) (x−(2/3)y)3
Using identity, (x –y)3 = x3–y3–3xy(x–y)
\((x - \frac{2}{3}y)^3\) = \((x)^3 -(\frac{2}{3}y)^3\) - \((3 \times x \times \frac{2}{3}y)(x - \frac{2}{3}y)\)
= \((x)^3 - \frac{8}{27}y^3 - 2xy(x - \frac{2}{3}y)\)
= \((x)^3 - \frac{8}{27}y^3 - 2x^2y + \frac{4}{3}xy^2\)