Answer
$$(x^2-3y)^4=x^8-12x^6y+54x^4y^2-108x^2y^3+81y^4$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}(x^2-3y)^4& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}x^8-12x^6y+54x^4y^2-108x^2y^3+81y^4\end{aligned} $$ | |
| ① | $$ (x^2-3y)^4 = (x^2-3y)^2 \cdot (x^2-3y)^2 $$ |
| ② | Find $ \left(x^2-3y\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}{ 3y }$. $$ \begin{aligned}\left(x^2-3y\right)^2 = \color{blue}{\left( x^2 \right)^2} -2 \cdot x^2 \cdot 3y + \color{red}{\left( 3y \right)^2} = x^4-6x^2y+9y^2\end{aligned} $$ |
| ③ | Multiply each term of $ \left( \color{blue}{x^4-6x^2y+9y^2}\right) $ by each term in $ \left( x^4-6x^2y+9y^2\right) $. $$ \left( \color{blue}{x^4-6x^2y+9y^2}\right) \cdot \left( x^4-6x^2y+9y^2\right) = \\ = x^8-6x^6y+9x^4y^2-6x^6y+36x^4y^2-54x^2y^3+9x^4y^2-54x^2y^3+81y^4 $$ |
| ④ | Combine like terms: $$ x^8 \color{blue}{-6x^6y} + \color{red}{9x^4y^2} \color{blue}{-6x^6y} + \color{green}{36x^4y^2} \color{orange}{-54x^2y^3} + \color{green}{9x^4y^2} \color{orange}{-54x^2y^3} +81y^4 = \\ = x^8 \color{blue}{-12x^6y} + \color{green}{54x^4y^2} \color{orange}{-108x^2y^3} +81y^4 $$ |
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