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\((-i+\sqrt{3})^{300}+(-i-\sqrt{3})^{300}=\)
Options:
Solution:
1195 Upvotes
Verified Answer
The correct answer is:
\(2^{301}\)
\(\begin{aligned}
&(-i+\sqrt{3})^{300}+(-i-\sqrt{3})^{300}=(i-\sqrt{3})^{300} +(-i-\sqrt{3})^{300} \\
&= i^{300}(1+i \sqrt{3})^{300}+i^{300}(-1+i \sqrt{3})^{300} \\
&=\left(-2 w^2\right)^{300}+(2 w)^{300} \quad\{\text {where } w \text { is cube root of unity}\} \\
&= 2^{300}\left[\left(w^2\right)^{300}+w^{300}\right]=2^{300} \times 2=2^{301}
\end{aligned}\)
Hence, option (b) is correct.
&(-i+\sqrt{3})^{300}+(-i-\sqrt{3})^{300}=(i-\sqrt{3})^{300} +(-i-\sqrt{3})^{300} \\
&= i^{300}(1+i \sqrt{3})^{300}+i^{300}(-1+i \sqrt{3})^{300} \\
&=\left(-2 w^2\right)^{300}+(2 w)^{300} \quad\{\text {where } w \text { is cube root of unity}\} \\
&= 2^{300}\left[\left(w^2\right)^{300}+w^{300}\right]=2^{300} \times 2=2^{301}
\end{aligned}\)
Hence, option (b) is correct.
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