Question

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

Answer 1

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

Use the formula ${\left(u+v\right)}^3=u^3+v^3+3u^{2}v+3u{v}^2$.

Put $u=\frac{3}{2}x,v=1$.

$\begin{array}{rcl}{\left(\frac{3}{2}x+1\right)}^3& =& {\left(\frac{3}{2}x\right)}^3+1^3+3{\left(\frac{3}{2}x\right)}^2\left(1\right)+3\left(\frac{3}{2}x\right){\left(1\right)}^2\\ \therefore {\left(\frac{3}{2}x+1\right)}^3& =& \frac{27}{8}{x}^3+1+\frac{27}{4}{x}^2+\frac{9}{2}x\end{array}$

Hence, expanded form of ${\left(\frac{3}{2}x+1\right)}^3$ is $\frac{27}{8}{x}^3+1+\frac{27}{4}{x}^2+\frac{9}{2}x$.