Answer
$$(2x^3-y^4)^2=y^8-4x^3y^4+4x^6$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}(2x^3-y^4)^2& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}4x^6-4x^3y^4+y^8 \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}y^8-4x^3y^4+4x^6\end{aligned} $$ | |
| ① | Find $ \left(2x^3-y^4\right)^2 $ using formula. $$ (A - B)^2 = \color{blue}{A^2} - 2 \cdot A \cdot B + \color{red}{B^2} $$where $ A = \color{blue}{ 2x^3 } $ and $ B = \color{red}{ y^4 }$. $$ \begin{aligned}\left(2x^3-y^4\right)^2 = \color{blue}{\left( 2x^3 \right)^2} -2 \cdot 2x^3 \cdot y^4 + \color{red}{\left( y^4 \right)^2} = 4x^6-4x^3y^4+y^8\end{aligned} $$ |
| ② | Combine like terms: $$ y^8-4x^3y^4+4x^6 = y^8-4x^3y^4+4x^6 $$ |
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