Question
Answer
$$(4x^4-y)^2=16x^8-8x^4y+y^2$$
Explanation
| $$ \begin{aligned}(4x^4-y)^2& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}16x^8-8x^4y+y^2\end{aligned} $$ | |
| ① | Find $ \left(4x^4-y\right)^2 $ using formula. $$ (A - B)^2 = \color{blue}{A^2} - 2 \cdot A \cdot B + \color{red}{B^2} $$where $ A = \color{blue}{ 4x^4 } $ and $ B = \color{red}{ y }$. $$ \begin{aligned}\left(4x^4-y\right)^2 = \color{blue}{\left( 4x^4 \right)^2} -2 \cdot 4x^4 \cdot y + \color{red}{y^2} = 16x^8-8x^4y+y^2\end{aligned} $$ |
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