## Precalculus (6th Edition)

$$11y^{3}-18y^{2}-4y$$
Initial Equation: $-y(y^{2}-4)+6y^{2}(2y-3)$ Distribute the $-y$ and $6y^{2}$ to their respective parenthesis: $$-y^{3}-4y+12y^{3}-18y^{2}$$ Then, combine like terms to arrive at final answer: $$11y^{3}-18y^{2}-4y$$