Answer
$$3x(2x^2+xy-y^2)-y(4x^2-xy+3y^2)=6x^3-x^2y-2xy^2-3y^3$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}3x(2x^2+xy-y^2)-y(4x^2-xy+3y^2)& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}6x^3+3x^2y-3xy^2-(4x^2y-xy^2+3y^3) \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}6x^3+3x^2y-3xy^2-4x^2y+xy^2-3y^3 \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}6x^3-x^2y-2xy^2-3y^3\end{aligned} $$ | |
| ① | Multiply $ \color{blue}{3x} $ by $ \left( 2x^2+xy-y^2\right) $ $$ \color{blue}{3x} \cdot \left( 2x^2+xy-y^2\right) = 6x^3+3x^2y-3xy^2 $$Multiply $ \color{blue}{y} $ by $ \left( 4x^2-xy+3y^2\right) $ $$ \color{blue}{y} \cdot \left( 4x^2-xy+3y^2\right) = 4x^2y-xy^2+3y^3 $$ |
| ② | Remove the parentheses by changing the sign of each term within them. $$ - \left( 4x^2y-xy^2+3y^3 \right) = -4x^2y+xy^2-3y^3 $$ |
| ③ | Combine like terms: $$ 6x^3+ \color{blue}{3x^2y} \color{red}{-3xy^2} \color{blue}{-4x^2y} + \color{red}{xy^2} -3y^3 = 6x^3 \color{blue}{-x^2y} \color{red}{-2xy^2} -3y^3 $$ |
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