Question
Answer
$$(m+n)((m^2+n^2)^2+2m^2n(m-n))=m^5+3m^4n+2m^3n^2+mn^4+n^5$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}(m+n)((m^2+n^2)^2+2m^2n(m-n))& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}(m+n)(1m^4+2m^2n^2+n^4+2m^2n(m-n)) \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}(m+n)(1m^4+2m^2n^2+n^4+2m^3n-2m^2n^2) \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}(m+n)(1m^4+2m^3n+n^4) \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}m^5+2m^4n+mn^4+m^4n+2m^3n^2+n^5 \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle5}{\textcircled {5}} } }}}m^5+3m^4n+2m^3n^2+mn^4+n^5\end{aligned} $$ | |
| ① | Find $ \left(m^2+n^2\right)^2 $ using formula. $$ (A + B)^2 = \color{blue}{A^2} + 2 \cdot A \cdot B + \color{red}{B^2} $$where $ A = \color{blue}{ m^2 } $ and $ B = \color{red}{ n^2 }$. $$ \begin{aligned}\left(m^2+n^2\right)^2 = \color{blue}{\left( m^2 \right)^2} +2 \cdot m^2 \cdot n^2 + \color{red}{\left( n^2 \right)^2} = m^4+2m^2n^2+n^4\end{aligned} $$ |
| ② | Multiply $ \color{blue}{2m^2n} $ by $ \left( m-n\right) $ $$ \color{blue}{2m^2n} \cdot \left( m-n\right) = 2m^3n-2m^2n^2 $$ |
| ③ | Combine like terms: $$ m^4+ \, \color{blue}{ \cancel{2m^2n^2}} \,+n^4+2m^3n \, \color{blue}{ -\cancel{2m^2n^2}} \, = m^4+2m^3n+n^4 $$ |
| ④ | Multiply each term of $ \left( \color{blue}{m+n}\right) $ by each term in $ \left( m^4+2m^3n+n^4\right) $. $$ \left( \color{blue}{m+n}\right) \cdot \left( m^4+2m^3n+n^4\right) = m^5+2m^4n+mn^4+m^4n+2m^3n^2+n^5 $$ |
| ⑤ | Combine like terms: $$ m^5+ \color{blue}{2m^4n} +mn^4+ \color{blue}{m^4n} +2m^3n^2+n^5 = m^5+ \color{blue}{3m^4n} +2m^3n^2+mn^4+n^5 $$ |
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