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

To expand the expression $\left[\frac{3}{2} x+1\right]^{3}$,we use the algebraic identity $(a+b)^{3} = a^{3} + b^{3} + 3ab(a+b)$.
Here,$a = \frac{3}{2}x$ and $b = 1$.
Substituting these values into the identity:
$\left[\frac{3}{2} x+1\right]^{3} = \left(\frac{3}{2} x\right)^{3} + (1)^{3} + 3\left(\frac{3}{2} x\right)(1)\left(\frac{3}{2} x + 1\right)$
Calculating the powers and the product:
$= \frac{27}{8} x^{3} + 1 + \frac{9}{2} x \left(\frac{3}{2} x + 1\right)$
Distributing $\frac{9}{2}x$ inside the bracket:
$= \frac{27}{8} x^{3} + 1 + \left(\frac{9}{2} x \cdot \frac{3}{2} x\right) + \left(\frac{9}{2} x \cdot 1\right)$
$= \frac{27}{8} x^{3} + 1 + \frac{27}{4} x^{2} + \frac{9}{2} x$
Rearranging the terms in descending order of powers:
$= \frac{27}{8} x^{3} + \frac{27}{4} x^{2} + \frac{9}{2} x + 1$