Question
Answer
$$(20-13w^2)^2+(32w-w^3)^2=w^6+105w^4+504w^2+400$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}(20-13w^2)^2+(32w-w^3)^2& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}400-520w^2+169w^4+1024w^2-64w^4+w^6 \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}w^6+105w^4+504w^2+400\end{aligned} $$ | |
| ① | Find $ \left(20-13w^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}{ 20 } $ and $ B = \color{red}{ 13w^2 }$. $$ \begin{aligned}\left(20-13w^2\right)^2 = \color{blue}{20^2} -2 \cdot 20 \cdot 13w^2 + \color{red}{\left( 13w^2 \right)^2} = 400-520w^2+169w^4\end{aligned} $$Find $ \left(32w-w^3\right)^2 $ using formula. $$ (A - B)^2 = \color{blue}{A^2} - 2 \cdot A \cdot B + \color{red}{B^2} $$where $ A = \color{blue}{ 32w } $ and $ B = \color{red}{ w^3 }$. $$ \begin{aligned}\left(32w-w^3\right)^2 = \color{blue}{\left( 32w \right)^2} -2 \cdot 32w \cdot w^3 + \color{red}{\left( w^3 \right)^2} = 1024w^2-64w^4+w^6\end{aligned} $$ |
| ② | Combine like terms: $$ 400 \color{blue}{-520w^2} + \color{red}{169w^4} + \color{blue}{1024w^2} \color{red}{-64w^4} +w^6 = w^6+ \color{red}{105w^4} + \color{blue}{504w^2} +400 $$ |
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