Question

Write the following cube 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 ${\left(A+B\right)}^{\mathbf{3}}\mathbf{=}{\mathbf{A}}^{\mathbf{3}}\mathbf{+}{\mathbf{B}}^{\mathbf{3}}\mathbf{+}\mathbf{3}{\mathbf{A}}^{\mathbf{2}}\mathbf{B}\mathbf{+}\mathbf{3}{\mathbf{AB}}^{\mathbf{2}}$.

Put $A=\frac{3}{2}X,B=1$.

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

Hence, the expanded form is $\begin{array}{rcl}{\left(\frac{3}{2}X+1\right)}^{\mathbf{3}}& \mathbf{=}& \frac{\mathbf{27}}{\mathbf{8}}{\mathbf{X}}^{\mathbf{3}}\mathbf{+}\mathbf{1}\mathbf{+}\frac{\mathbf{27}}{\mathbf{4}}{\mathbf{X}}^{\mathbf{2}}\mathbf{+}\frac{\mathbf{9}}{\mathbf{2}}\mathbf{X}\end{array}$ .