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