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