Answer
$$(x^2-2y^3)^4=16y^{12}-32x^2y^9+24x^4y^6-8x^6y^3+x^8$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}(x^2-2y^3)^4& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}16y^{12}-32x^2y^9+24x^4y^6-8x^6y^3+x^8\end{aligned} $$ | |
| ① | $$ (x^2-2y^3)^4 = (x^2-2y^3)^2 \cdot (x^2-2y^3)^2 $$ |
| ② | Find $ \left(x^2-2y^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}{ x^2 } $ and $ B = \color{red}{ 2y^3 }$. $$ \begin{aligned}\left(x^2-2y^3\right)^2 = \color{blue}{\left( x^2 \right)^2} -2 \cdot x^2 \cdot 2y^3 + \color{red}{\left( 2y^3 \right)^2} = x^4-4x^2y^3+4y^6\end{aligned} $$ |
| ③ | Multiply each term of $ \left( \color{blue}{x^4-4x^2y^3+4y^6}\right) $ by each term in $ \left( x^4-4x^2y^3+4y^6\right) $. $$ \left( \color{blue}{x^4-4x^2y^3+4y^6}\right) \cdot \left( x^4-4x^2y^3+4y^6\right) = \\ = x^8-4x^6y^3+4x^4y^6-4x^6y^3+16x^4y^6-16x^2y^9+4x^4y^6-16x^2y^9+16y^{12} $$ |
| ④ | Combine like terms: $$ x^8 \color{blue}{-4x^6y^3} + \color{red}{4x^4y^6} \color{blue}{-4x^6y^3} + \color{green}{16x^4y^6} \color{orange}{-16x^2y^9} + \color{green}{4x^4y^6} \color{orange}{-16x^2y^9} +16y^{12} = \\ = 16y^{12} \color{orange}{-32x^2y^9} + \color{green}{24x^4y^6} \color{blue}{-8x^6y^3} +x^8 $$ |
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