Question

Write the following cubes in expanded from:
(i) $(2x+1{)}^3$ (ii) $(2a-3b{)}^3$ (iii) ${\left[\frac{3}{2}x+1\right]}^3$ (iv) ${[x-\frac{2}{3}y]}^3$

Answer 1

Since, we have $(a+b{)}^3=a^3+b^3+3ab(a+b)$.....(i)
$(a-b{)}^3=a^3-b^3-3ab(a-b)$......(ii)
(i)Given $(2x+1{)}^3=(2x{)}^3+(1{)}^3+3(2x\left)\right(1\left)\right(2x+1)$ [Substitute $a=2x$ and $b=1$ in equation (i)]
$=8x^3+1+6x(2x+1)=8x^3+12x^2+6x+1$

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

$=8a^3-27b^3-18ab(2a-3b)$ [ from equation(ii)]
$=8a^3-27b^3-36a^{2}b+54a{b}^2$

(iii) ${(\frac{3}{2}x+1)}^3=\left(\frac{3}{2}\right)x^3+(1{)}^3+3\left(\frac{3}{2}x\right)(1)(\frac{3}{2}x+1)$ [from equation(i)]

$=\frac{27}{8}{x}^3+1+\frac{9x}{2}(\frac{3x}{2}+1)=\frac{27}{8}{x}^3+\frac{27}{4}{x}^2+\frac{9}{2}x+1$

(iv) ${[x-\frac{2}{3}y]}^3=x^3-(\frac{2}{3}y{)}^3-3x(\frac{2}{3}y\left)\right(x-\frac{2}{3}y)$ [From equation(ii)]
= $x^3-(\frac{2}{3}{)}^3\times y^3-3\times x^2\times \frac{2}{3}y+3\times x\times (\frac{2}{3}y{)}^2$
$=x^3-\frac{8}{27}{y}^3-2x^{2}y+\frac{4}{3}x{y}^2$