Algebra 1

$\frac{(y^{2})(y-1)}{y-3}$
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.