Answer
$$(1-y)^3y=-y^4+3y^3-3y^2+y$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}(1-y)^3y& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}(1-3y+3y^2-y^3)y \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}y-3y^2+3y^3-y^4 \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}-y^4+3y^3-3y^2+y\end{aligned} $$ | |
| ① | Find $ \left(1-y\right)^3 $ using formula $$ (A - B) = A^3 - 3A^2B + 3AB^2 - B^3 $$where $ A = 1 $ and $ B = y $. $$ \left(1-y\right)^3 = 1^3-3 \cdot 1^2 \cdot y + 3 \cdot 1 \cdot y^2-y^3 = 1-3y+3y^2-y^3 $$ |
| ② | $$ \left( \color{blue}{1-3y+3y^2-y^3}\right) \cdot y = y-3y^2+3y^3-y^4 $$ |
| ③ | Combine like terms: $$ -y^4+3y^3-3y^2+y = -y^4+3y^3-3y^2+y $$ |
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