Answer
$$8(x+1)^3(16x-7)+36=128x^4+328x^3+216x^2-40x-20$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}8(x+1)^3(16x-7)+36& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}8(x^3+3x^2+3x+1)(16x-7)+36 \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}(8x^3+24x^2+24x+8)(16x-7)+36 \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}128x^4+328x^3+216x^2-40x-56+36 \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle5}{\textcircled {5}} } }}}128x^4+328x^3+216x^2-40x-20\end{aligned} $$ | |
| ① | Find $ \left(x+1\right)^3 $ using formula $$ (A + B) = A^3 + 3A^2B + 3AB^2 + B^3 $$where $ A = x $ and $ B = 1 $. $$ \left(x+1\right)^3 = x^3+3 \cdot x^2 \cdot 1 + 3 \cdot x \cdot 1^2+1^3 = x^3+3x^2+3x+1 $$ |
| ② | Multiply $ \color{blue}{8} $ by $ \left( x^3+3x^2+3x+1\right) $ $$ \color{blue}{8} \cdot \left( x^3+3x^2+3x+1\right) = 8x^3+24x^2+24x+8 $$ |
| ③ | Multiply each term of $ \left( \color{blue}{8x^3+24x^2+24x+8}\right) $ by each term in $ \left( 16x-7\right) $. $$ \left( \color{blue}{8x^3+24x^2+24x+8}\right) \cdot \left( 16x-7\right) = 128x^4-56x^3+384x^3-168x^2+384x^2-168x+128x-56 $$ |
| ④ | Combine like terms: $$ 128x^4 \color{blue}{-56x^3} + \color{blue}{384x^3} \color{red}{-168x^2} + \color{red}{384x^2} \color{green}{-168x} + \color{green}{128x} -56 = \\ = 128x^4+ \color{blue}{328x^3} + \color{red}{216x^2} \color{green}{-40x} -56 $$ |
| ⑤ | Combine like terms: $$ 128x^4+328x^3+216x^2-40x \color{blue}{-56} + \color{blue}{36} = 128x^4+328x^3+216x^2-40x \color{blue}{-20} $$ |
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