Answer
$$(4-ix)^4=i^4x^4-16i^3x^3+96i^2x^2-256ix+256$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}(4-ix)^4& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}i^4x^4-16i^3x^3+96i^2x^2-256ix+256\end{aligned} $$ | |
| ① | $$ (4-ix)^4 = (4-ix)^2 \cdot (4-ix)^2 $$ |
| ② | Find $ \left(4-ix\right)^2 $ using formula. $$ (A - B)^2 = \color{blue}{A^2} - 2 \cdot A \cdot B + \color{red}{B^2} $$where $ A = \color{blue}{ 4 } $ and $ B = \color{red}{ ix }$. $$ \begin{aligned}\left(4-ix\right)^2 = \color{blue}{4^2} -2 \cdot 4 \cdot ix + \color{red}{\left( ix \right)^2} = 16-8ix+i^2x^2\end{aligned} $$ |
| ③ | Multiply each term of $ \left( \color{blue}{16-8ix+i^2x^2}\right) $ by each term in $ \left( 16-8ix+i^2x^2\right) $. $$ \left( \color{blue}{16-8ix+i^2x^2}\right) \cdot \left( 16-8ix+i^2x^2\right) = \\ = 256-128ix+16i^2x^2-128ix+64i^2x^2-8i^3x^3+16i^2x^2-8i^3x^3+i^4x^4 $$ |
| ④ | Combine like terms: $$ 256 \color{blue}{-128ix} + \color{red}{16i^2x^2} \color{blue}{-128ix} + \color{green}{64i^2x^2} \color{orange}{-8i^3x^3} + \color{green}{16i^2x^2} \color{orange}{-8i^3x^3} +i^4x^4 = \\ = i^4x^4 \color{orange}{-16i^3x^3} + \color{green}{96i^2x^2} \color{blue}{-256ix} +256 $$ |
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