Question

Write the following cubes in expanded form:

(i) $(2x+1{)}^3$

(ii)$(2a-3b{)}^3$

(iii) $(\frac{3}{2}x+1{)}^3$

(iv) $(x-\frac{2}{3}y{)}^3$

Answer 1

It is known that,

$(a+b{)}^3=a^3+b^3+3ab(a+b)$...(A) and

$(a-b{)}^3=a^3-b^3-3ab(a-b)$...(B)

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

$=8x^3+1+6x(2x+1)$

$=8x^3+1+12x^2+6x$

$=8x^3+12x^2+6x+1$

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

$=8a^3-27b^3-18ab(2a-3b)$

$=8a^3-27b^3-36a^{2}b+54a{b}^2$

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

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

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

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

(vi) $(x-\frac{2}{3}y{)}^3=x^3-(\frac{2}{3}y{)}^3-3x(x-\frac{2}{3}y)$

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

$=x^3-\frac{8}{27}{y}^3-2x^{2}y+\frac{4}{3}x{y}^2$