Answer
$\color{blue}{3(3r+1)^{-2/3}(3r^2+3r+1)}$
Work Step by Step
The least exponent of the binomials is $-2/3$.
Factor out $(3r+1)^{-2/3}$ to obtain:
$=(3r+1)^{-2/3}[1+(3r+1)+(3r+1)^{2}]
\\=(3r+1)^{-2/3}[3r+2+(3r+1)^{2}]$
Square the binomial then combine like terms to obtain:
$=(3r+1)^{-2/3}(3r+2+9r^2+6r+1)
\\=(3r+1)^{-2/3}(9r^2+9r+3)$
Factor out $3$ in the trinomial to obtain:
$\\=\color{blue}{3(3r+1)^{-2/3}(3r^2+3r+1)}$
