Question
Answer
$$4(n+1)^3-(n+1)=4n^3+12n^2+11n+3$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}4(n+1)^3-(n+1)& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}4(1n^3+3n^2+3n+1)-(n+1) \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}4n^3+12n^2+12n+4-(n+1) \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}4n^3+12n^2+12n+4-n-1 \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}4n^3+12n^2+11n+3\end{aligned} $$ | |
| 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 $ \color{blue}{4} $ by $ \left( n^3+3n^2+3n+1\right) $ $$ \color{blue}{4} \cdot \left( n^3+3n^2+3n+1\right) = 4n^3+12n^2+12n+4 $$ |
| ③ | Remove the parentheses by changing the sign of each term within them. $$ - \left( n+1 \right) = -n-1 $$ |
| ④ | Combine like terms: $$ 4n^3+12n^2+ \color{blue}{12n} + \color{red}{4} \color{blue}{-n} \color{red}{-1} = 4n^3+12n^2+ \color{blue}{11n} + \color{red}{3} $$ |
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