Write the following cubes in expanded form : $\left[\frac{3}{2} x+1\right]^{3}$

Using Identity $VI$ and Identity $VII,$ we have

$(x+y)^{3}=x^{3}+y^{3}+3 x y(x+y),$ and $(x-y)^{3}=x^{3}-y^{3}-3 x y(x-y)$

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

$=\frac{27}{8} x^{3}+1+\frac{9}{2} x\left[\frac{3}{2} x+1\right]$             [Using Identity $VI$]

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