Question
Answer
$$(n^3-m)^4=n^{12}-4mn^9+6m^2n^6-4m^3n^3+m^4$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}(n^3-m)^4& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}n^{12}-4mn^9+6m^2n^6-4m^3n^3+m^4\end{aligned} $$ | |
| ① | $$ (n^3-m)^4 = (n^3-m)^2 \cdot (n^3-m)^2 $$ |
| ② | Find $ \left(n^3-m\right)^2 $ using formula. $$ (A - B)^2 = \color{blue}{A^2} - 2 \cdot A \cdot B + \color{red}{B^2} $$where $ A = \color{blue}{ n^3 } $ and $ B = \color{red}{ m }$. $$ \begin{aligned}\left(n^3-m\right)^2 = \color{blue}{\left( n^3 \right)^2} -2 \cdot n^3 \cdot m + \color{red}{m^2} = n^6-2mn^3+m^2\end{aligned} $$ |
| ③ | Multiply each term of $ \left( \color{blue}{n^6-2mn^3+m^2}\right) $ by each term in $ \left( n^6-2mn^3+m^2\right) $. $$ \left( \color{blue}{n^6-2mn^3+m^2}\right) \cdot \left( n^6-2mn^3+m^2\right) = \\ = n^{12}-2mn^9+m^2n^6-2mn^9+4m^2n^6-2m^3n^3+m^2n^6-2m^3n^3+m^4 $$ |
| ④ | Combine like terms: $$ n^{12} \color{blue}{-2mn^9} + \color{red}{m^2n^6} \color{blue}{-2mn^9} + \color{green}{4m^2n^6} \color{orange}{-2m^3n^3} + \color{green}{m^2n^6} \color{orange}{-2m^3n^3} +m^4 = \\ = n^{12} \color{blue}{-4mn^9} + \color{green}{6m^2n^6} \color{orange}{-4m^3n^3} +m^4 $$ |
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