Question

Evaluate the following : ${\left(\frac{3}{2}x+1\right)}^3-\frac{27x^2}{4}-\frac{9x}{2}.$

Answer 1

Using the identity (a + b)³ = a³ + b³ + 3ab(a + b), we get
$$\begin{gathered} {\left(\frac{3}{2}x+1\right)}^3-\frac{27x^2}{4}-\frac{9x}{2} \\ =\left[{\left(\frac{3x}{2}\right)}^3+1^3+3\times \frac{3x}{2}\times 1\left(\frac{3x}{2}+1\right)\right]-\frac{27x^2}{4}-\frac{9x}{2} \\ =\frac{27x^3}{8}+1+\frac{27x^2}{4}+\frac{9x}{2}-\frac{27x^2}{4}-\frac{9x}{2} \\ =\frac{27x^3}{8}+1 \end{gathered}$$
Hence, ${\left(\frac{3}{2}x+1\right)}^3-\frac{27x^2}{4}-\frac{9x}{2}=\frac{27x^3}{8}+1$.