Answer
$\frac{(y^{2})(y-1)}{y-3}$
Work Step by Step
Simplifying the question,
$\frac{(y^{4}-y^{2})}{y^2-2y-3}$
Factorising the numerator,
=$\frac{(y^{4}-y^{2})}{y^2-2y-3}$ = $\frac{y^{2}(y^{2}-1)}{y^2-2y-3}$ = $\frac{y^{2}(y-1)(y+1)}{y^2-2y-3}$
Factorising the denominator,
=$\frac{y^{2}(y-1)(y+1)}{y^2-2y-3}$=$\frac{y^{2}(y-1)(y+1)}{y^2-3y+y-3}$
=$\frac{y^{2}(y-1)(y+1)}{y(y-3)+1(y-3)}$
=$\frac{y^{2}(y-1)(y+1)}{(y+1)(y-3)}$
Cancel the common terms,
=$\frac{y^{2}(y-1)}{(y-3)}$
This is the answer.
