(3y^3+y^2-6y)cdot (-(1)/(3)y^2)

To solve the expression \((3y^{3}+y^{2}-6y) \cdot (-\frac{1}{3}y^{2})\), we need to distribute \(-\frac{1}{3}y^{2}\) to each term inside the parentheses.<br /><br />Let's break it down step by step:<br /><br />1. Distribute \(-\frac{1}{3}y^{2}\) to \(3y^{3}\):<br /> \[<br /> 3y^{3} \cdot (-\frac{1}{3}y^{2}) = 3 \cdot (-\frac{1}{3}) \cdot y^{3} \cdot y^{2} = -y^{5}<br /> \]<br /><br />2. Distribute \(-\frac{1}{3}y^{2}\) to \(y^{2}\):<br /> \[<br /> y^{2} \cdot (-\frac{1}{3}y^{2}) = 1 \cdot (-\frac{1}{3}) \cdot y^{2} \cdot y^{2} = -\frac{1}{3}y^{4}<br /> \]<br /><br />3. Distribute \(-\frac{1}{3}y^{2}\) to \(-6y\):<br /> \[<br /> -6y \cdot (-\frac{1}{3}y^{2}) = -6 \cdot (-\frac{1}{3}) \cdot y \cdot y^{2} = 2y^{3}<br /> \]<br /><br />Now, combine all the terms:<br />\[<br />- y^{5} - \frac{1}{3}y^{4} + 2y^{3}<br />\]<br /><br />So, the final answer is:<br />\[<br />(3y^{3}+y^{2}-6y) \cdot (-\frac{1}{3}y^{2}) = -y^{5} - \frac{1}{3}y^{4} + 2y^{3}<br />\]