Question

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

Answer 1

$\left(i\right)(2x+1{)}^3$ [Using identity $(a+b{)}^3=a^3+b^3+3ab(a+b)$]
$=(2x{)}^3+1^3+3(2x\left)\right(1\left)\right(2x+1)$
$=8x^3+1+6x(2x+1)$
$=8x^3+12x^2+6x+1$

$\left(ii\right)(2a-3b{)}^3$ [Using identity $(a-b{)}^3=a^3-b^3-3ab(a-b)$

$=(2a{)}^3-(3b{)}^3-3\times 2a\times 3b(2a-3b)$
$=8a^3-27b^3-18ab(2a-3b)$
$=8a^3-27b^3-36a^{2}b+54a{b}^2$

$\left(iii\right){[\frac{3}{2}x+1]}^3$ [Using identity $(a+b{)}^3=a^3+b^3+3ab(a+b)$

$={\left(\frac{3}{2}x\right)}^3+1^3+3\left(\frac{3}{2}x\right)\left(1\right)(\frac{3}{2}x+1)$

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

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

$\left(iv\right){[x-\frac{2}{3}y]}^3$ [Using identity $(a-b{)}^3=a^3-b^3-3ab(a-b)$]

$=x^3-{\left(\frac{2}{3}y\right)}^3-3\left(x\right)\left(\frac{2}{3}y\right)(x-\frac{2}{3}y)$
$=x^3-\frac{8}{27}{y}^3-2xy(x-\frac{2}{3}y)$
$=x^3-\frac{8}{27}{y}^3-2x^{2}y+\frac{4}{3}x{y}^2$