Question
Answer
$$(n^3+9)(n+1)^3+9(n^3+9)=n^6+3n^5+3n^4+19n^3+27n^2+27n+90$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}(n^3+9)(n+1)^3+9(n^3+9)& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}(n^3+9)(1n^3+3n^2+3n+1)+9(n^3+9) \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}n^6+3n^5+3n^4+10n^3+27n^2+27n+9+9n^3+81 \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}n^6+3n^5+3n^4+19n^3+27n^2+27n+90\end{aligned} $$ | |
| ① | Find $ \left(n+1\right)^3 $ using formula $$ (A + B) = A^3 + 3A^2B + 3AB^2 + B^3 $$where $ A = n $ and $ B = 1 $. $$ \left(n+1\right)^3 = n^3+3 \cdot n^2 \cdot 1 + 3 \cdot n \cdot 1^2+1^3 = n^3+3n^2+3n+1 $$ |
| ② | Multiply each term of $ \left( \color{blue}{n^3+9}\right) $ by each term in $ \left( n^3+3n^2+3n+1\right) $. $$ \left( \color{blue}{n^3+9}\right) \cdot \left( n^3+3n^2+3n+1\right) = n^6+3n^5+3n^4+n^3+9n^3+27n^2+27n+9 $$ |
| ③ | Combine like terms: $$ n^6+3n^5+3n^4+ \color{blue}{n^3} + \color{blue}{9n^3} +27n^2+27n+9 = n^6+3n^5+3n^4+ \color{blue}{10n^3} +27n^2+27n+9 $$Multiply $ \color{blue}{9} $ by $ \left( n^3+9\right) $ $$ \color{blue}{9} \cdot \left( n^3+9\right) = 9n^3+81 $$ |
| ④ | Combine like terms: $$ n^6+3n^5+3n^4+ \color{blue}{10n^3} +27n^2+27n+ \color{red}{9} + \color{blue}{9n^3} + \color{red}{81} = n^6+3n^5+3n^4+ \color{blue}{19n^3} +27n^2+27n+ \color{red}{90} $$ |
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