Answer
$\frac{3\left( y+1 \right)\left( y-1 \right)}{y}$
Work Step by Step
$\frac{{{y}^{3}}-y}{3y+1}\div \frac{{{y}^{2}}}{9y+3}$
Simplify the expression as,
$\begin{align}
& \frac{{{y}^{3}}-y}{3y+1}\div \frac{{{y}^{2}}}{9y+3}=\frac{y\left( {{y}^{2}}-{{1}^{2}} \right)}{3y+1}\cdot \frac{3\left( 3y+1 \right)}{{{y}^{2}}} \\
& =y\left( y-1 \right)\left( y+1 \right)\cdot \frac{3}{{{y}^{2}}} \\
& =\frac{3\left( y+1 \right)\left( y-1 \right)}{y}
\end{align}$
Thus,
$\frac{{{y}^{3}}-y}{3y+1}\div \frac{{{y}^{2}}}{9y+3}=\frac{3\left( y+1 \right)\left( y-1 \right)}{y}$
Thus, $\frac{{{y}^{3}}-y}{3y+1}\div \frac{{{y}^{2}}}{9y+3}$ reduces to $\frac{3\left( y+1 \right)\left( y-1 \right)}{y}$.
