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If $\mathrm{n}$ is an integer which leaves remainder one when divided by three, then $(1+\sqrt{3} i)^{n}+(1-\sqrt{3} i)^{n}$ equals
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Verified Answer
The correct answer is:
$-(-2)^{\mathrm{n}}$
$$
\begin{aligned}
&(1+\sqrt{3} i)^{n}+(1-\sqrt{3} i)^{n} \\
=&\left[2\left(\frac{1+\sqrt{3} i}{2}\right)\right]^{n}+\left[2\left(\frac{1-\sqrt{3} i}{2}\right)\right]^{n} \\
=&\left(-2 \omega^{2}\right)^{n}+(-2 \omega)^{n} \\
=&(-2)^{n}\left[\left(\omega^{2}\right)^{3 r+1}+(\omega)^{3 r+1}\right] \\
&(\because n=3 r+1, \text { where } r \text { is an integer }) \\
=&(-2)^{n}\left(\omega^{2}+\omega\right)=-(-2)^{n}
\end{aligned}
$$
\begin{aligned}
&(1+\sqrt{3} i)^{n}+(1-\sqrt{3} i)^{n} \\
=&\left[2\left(\frac{1+\sqrt{3} i}{2}\right)\right]^{n}+\left[2\left(\frac{1-\sqrt{3} i}{2}\right)\right]^{n} \\
=&\left(-2 \omega^{2}\right)^{n}+(-2 \omega)^{n} \\
=&(-2)^{n}\left[\left(\omega^{2}\right)^{3 r+1}+(\omega)^{3 r+1}\right] \\
&(\because n=3 r+1, \text { where } r \text { is an integer }) \\
=&(-2)^{n}\left(\omega^{2}+\omega\right)=-(-2)^{n}
\end{aligned}
$$
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