Answer
$$(d-1)(d+2)(d+3)=d^3+4d^2+d-6$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}(d-1)(d+2)(d+3)& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}(1d^2+2d-d-2)(d+3) \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}(1d^2+d-2)(d+3) \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}d^3+3d^2+d^2+3d-2d-6 \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}d^3+4d^2+d-6\end{aligned} $$ | |
| ① | Multiply each term of $ \left( \color{blue}{d-1}\right) $ by each term in $ \left( d+2\right) $. $$ \left( \color{blue}{d-1}\right) \cdot \left( d+2\right) = d^2+2d-d-2 $$ |
| ② | Combine like terms: $$ d^2+ \color{blue}{2d} \color{blue}{-d} -2 = d^2+ \color{blue}{d} -2 $$ |
| ③ | Multiply each term of $ \left( \color{blue}{d^2+d-2}\right) $ by each term in $ \left( d+3\right) $. $$ \left( \color{blue}{d^2+d-2}\right) \cdot \left( d+3\right) = d^3+3d^2+d^2+3d-2d-6 $$ |
| ④ | Combine like terms: $$ d^3+ \color{blue}{3d^2} + \color{blue}{d^2} + \color{red}{3d} \color{red}{-2d} -6 = d^3+ \color{blue}{4d^2} + \color{red}{d} -6 $$ |
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