Question
Answer
$$1-(1-p^2)^3=p^6-3p^4+3p^2$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}1-(1-p^2)^3& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}1-(1-3p^2+3p^4-p^6) \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}1-1+3p^2-3p^4+p^6 \xlongequal{ } \\[1 em] & \xlongequal{ } \cancel{1} -\cancel{1}+3p^2-3p^4+p^6 \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}p^6-3p^4+3p^2\end{aligned} $$ | |
| ① | Find $ \left(1-p^2\right)^3 $ using formula $$ (A - B) = A^3 - 3A^2B + 3AB^2 - B^3 $$where $ A = 1 $ and $ B = p^2 $. $$ \left(1-p^2\right)^3 = 1^3-3 \cdot 1^2 \cdot p^2 + 3 \cdot 1 \cdot \left( p^2 \right)^2-\left( p^2 \right)^3 = 1-3p^2+3p^4-p^6 $$ |
| ② | Remove the parentheses by changing the sign of each term within them. $$ - \left( 1-3p^2+3p^4-p^6 \right) = -1+3p^2-3p^4+p^6 $$ |
| ③ | Combine like terms: $$ \, \color{blue}{ \cancel{1}} \, \, \color{blue}{ -\cancel{1}} \,+3p^2-3p^4+p^6 = p^6-3p^4+3p^2 $$ |
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