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If $2 \cos \theta=x+\frac{1}{x}$, then $2 \cos 3 \theta=$
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The correct answer is:
$x^3+\frac{1}{x^3}$
$\begin{aligned} & \text { We have } \cos \theta=\frac{1}{2}\left(x+\frac{1}{x}\right) \\ & 2 \cos 3 \theta \\ & =2\left[4 \cos ^3 \theta-3 \cos \theta\right]=2\left\{4\left[\left(\frac{1}{2}\right)\left(x+\frac{1}{x}\right)\right]^3-\left[3\left(\frac{1}{2}\right)\left(x+\frac{1}{x}\right)\right]\right\} \\ & =2\left[\left(\frac{1}{2}\right)\left(x^3+\frac{1}{x^3}\right)+3\left(x+\frac{1}{x}\right)\right]-3\left(x+\frac{1}{x}\right)=x^3+\frac{1}{x^3}\end{aligned}$
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