Answer
$$p\cdot(1-(1-p)^3)p=p^5-3p^4+3p^3$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}p\cdot(1-(1-p)^3)p& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}p\cdot(1-(1-3p+3p^2-p^3))p \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}p(1-1+3p-3p^2+p^3)p \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}p(1p^3-3p^2+3p)p \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}(1p^4-3p^3+3p^2)p \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle5}{\textcircled {5}} } }}}p^5-3p^4+3p^3\end{aligned} $$ | |
| ① | Find $ \left(1-p\right)^3 $ using formula $$ (A - B) = A^3 - 3A^2B + 3AB^2 - B^3 $$where $ A = 1 $ and $ B = p $. $$ \left(1-p\right)^3 = 1^3-3 \cdot 1^2 \cdot p + 3 \cdot 1 \cdot p^2-p^3 = 1-3p+3p^2-p^3 $$ |
| ② | Remove the parentheses by changing the sign of each term within them. $$ - \left( 1-3p+3p^2-p^3 \right) = -1+3p-3p^2+p^3 $$ |
| ③ | Combine like terms: $$ \, \color{blue}{ \cancel{1}} \, \, \color{blue}{ -\cancel{1}} \,+3p-3p^2+p^3 = p^3-3p^2+3p $$ |
| ④ | Multiply $ \color{blue}{p} $ by $ \left( p^3-3p^2+3p\right) $ $$ \color{blue}{p} \cdot \left( p^3-3p^2+3p\right) = p^4-3p^3+3p^2 $$ |
| ⑤ | $$ \left( \color{blue}{p^4-3p^3+3p^2}\right) \cdot p = p^5-3p^4+3p^3 $$ |
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