Question

Write the following cubes in expanded form:

($\frac{3x}{2}+1{)}^3$

Answer 1

Using identity ${(x+y)}^3$= $x^3+y^3+3xy(x+y)$

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

$=27\frac{\left(x^3\right)}{8}$ +1+ $\frac{9x}{2}$($\frac{3x}{2}$ + 1)

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

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