Answer
$$3x^3+x+2^3-4x^2+14y-2(y+x)=3x^3-4x^2-x+12y+8$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}3x^3+x+2^3-4x^2+14y-2(y+x)& \xlongequal{ }3x^3+x+8-4x^2+14y-2(y+x) \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}3x^3+x+8-4x^2+14y-(2y+2x) \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}3x^3+x+8-4x^2+14y-2y-2x \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}3x^3-4x^2-x+12y+8\end{aligned} $$ | |
| ① | Multiply $ \color{blue}{2} $ by $ \left( y+x\right) $ $$ \color{blue}{2} \cdot \left( y+x\right) = 2y+2x $$ |
| ② | Remove the parentheses by changing the sign of each term within them. $$ - \left( 2y+2x \right) = -2y-2x $$ |
| ③ | Combine like terms: $$ 3x^3+ \color{blue}{x} +8-4x^2+ \color{red}{14y} \color{red}{-2y} \color{blue}{-2x} = 3x^3-4x^2 \color{blue}{-x} + \color{red}{12y} +8 $$ |
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