• # question_answer If $x+\frac{1}{x}=2\,\cos \theta ,$ then ${{x}^{3}}+\frac{1}{{{x}^{3}}}=$ [MP PET 2004] A) $\cos \,\,3\theta$ B) $2\,\cos \,3\theta$ C) $\frac{1}{2}\cos \,3\theta$ D) $\frac{1}{3}\cos \,3\theta$

We have$x+\frac{1}{x}=2\cos \theta$, Now ${{x}^{3}}+\frac{1}{{{x}^{3}}}={{\left( x+\frac{1}{x} \right)}^{3}}-3x\frac{1}{x}\left( x+\frac{1}{x} \right)$ = ${{(2\cos \theta )}^{3}}-3(2\cos \theta )=8{{\cos }^{3}}\theta -6\cos \theta$ = $2(4{{\cos }^{3}}\theta -3\cos \theta )=2\cos 3\theta$. Trick: Put $x=1$ $\Rightarrow$ $\theta ={{0}^{{}^\circ }}$. Then${{x}^{3}}+\frac{1}{{{x}^{3}}}=2=2\cos 3\theta$.