Question
Answer
$$-3y^3-4y^3(1-y^3+x^2y)=-4x^2y^4+4y^6-7y^3$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}-3y^3-4y^3(1-y^3+x^2y)& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}-3y^3-(4y^3-4y^6+4x^2y^4) \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}-3y^3-4y^3+4y^6-4x^2y^4 \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}-4x^2y^4+4y^6-7y^3\end{aligned} $$ | |
| ① | Multiply $ \color{blue}{4y^3} $ by $ \left( 1-y^3+x^2y\right) $ $$ \color{blue}{4y^3} \cdot \left( 1-y^3+x^2y\right) = 4y^3-4y^6+4x^2y^4 $$ |
| ② | Remove the parentheses by changing the sign of each term within them. $$ - \left( 4y^3-4y^6+4x^2y^4 \right) = -4y^3+4y^6-4x^2y^4 $$ |
| ③ | Combine like terms: $$ \color{blue}{-3y^3} \color{blue}{-4y^3} +4y^6-4x^2y^4 = -4x^2y^4+4y^6 \color{blue}{-7y^3} $$ |
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