Question
Answer
$$(1-z^2)^4=z^8-4z^6+6z^4-4z^2+1$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}(1-z^2)^4& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}z^8-4z^6+6z^4-4z^2+1\end{aligned} $$ | |
| ① | $$ (1-z^2)^4 = (1-z^2)^2 \cdot (1-z^2)^2 $$ |
| ② | Find $ \left(1-z^2\right)^2 $ using formula. $$ (A - B)^2 = \color{blue}{A^2} - 2 \cdot A \cdot B + \color{red}{B^2} $$where $ A = \color{blue}{ 1 } $ and $ B = \color{red}{ z^2 }$. $$ \begin{aligned}\left(1-z^2\right)^2 = \color{blue}{1^2} -2 \cdot 1 \cdot z^2 + \color{red}{\left( z^2 \right)^2} = 1-2z^2+z^4\end{aligned} $$ |
| ③ | Multiply each term of $ \left( \color{blue}{1-2z^2+z^4}\right) $ by each term in $ \left( 1-2z^2+z^4\right) $. $$ \left( \color{blue}{1-2z^2+z^4}\right) \cdot \left( 1-2z^2+z^4\right) = 1-2z^2+z^4-2z^2+4z^4-2z^6+z^4-2z^6+z^8 $$ |
| ④ | Combine like terms: $$ 1 \color{blue}{-2z^2} + \color{red}{z^4} \color{blue}{-2z^2} + \color{green}{4z^4} \color{orange}{-2z^6} + \color{green}{z^4} \color{orange}{-2z^6} +z^8 = \\ = z^8 \color{orange}{-4z^6} + \color{green}{6z^4} \color{blue}{-4z^2} +1 $$ |
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