Question
Answer
$$(-sqrt\cdot3-i)^2=9q^2r^2s^2t^2+6iqrst+(-1)$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}(-sqrt\cdot3-i)^2& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}9q^2r^2s^2t^2+6iqrst+i^2 \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}9q^2r^2s^2t^2+6iqrst+(-1)\end{aligned} $$ | |
| ① | Find $ \left(-3qrst-i\right)^2 $ in two steps. S1: Change all signs inside bracket. S2: Apply formula $$ (A + B)^2 = \color{blue}{A^2} + 2 \cdot A \cdot B + \color{red}{B^2} $$where $ A = \color{blue}{ 3qrst } $ and $ B = \color{red}{ i }$. $$ \begin{aligned}\left(-3qrst-i\right)^2& \xlongequal{ S1 } \left(3qrst+i\right)^2 \xlongequal{ S2 } \color{blue}{\left( 3qrst \right)^2} +2 \cdot 3qrst \cdot i + \color{red}{i^2} = \\[1 em] & = 9q^2r^2s^2t^2+6iqrst+i^2\end{aligned} $$ |
| ② | $$ i^2 = -1 $$ |
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