If $x+\frac{1}{x}=2\cos \theta$, then $x^3+\frac{1}{x^3}=$

$2\cos \theta =x+\frac{1}{x}=\frac{x^2+1}{x}$

$\cos \theta =\frac{x^2+1}{2x}$

$2\cos 3\theta =2(4{\cos }^3\theta -3\cos \theta )$

$=8{\left(\frac{x^2+1}{2x}\right)}^3-6\left(\frac{x^2+1}{2x}\right)$

$=8\left(\frac{x^6+1+3x^4+3x^2}{8x^3}\right)-\frac{6(x^2+1)}{2x}$

$=x^3+\frac{1}{x^3}+3x+\frac{3}{x}-3x-\frac{3}{x}$

$=x^3+\frac{1}{x^3}$

(or)

$2\cos \theta =x+\frac{1}{x}$

$c.o.b.s$

$8{\cos }^3\theta =x^3+\frac{1}{x^3}+3.x.\frac{1}{x}(x+\frac{1}{x})$

$8{\cos }^3\theta =x^3+\frac{1}{x^3}+3\left(2\cos \theta \right)$

$8{\cos }^3\theta -6\cos \theta =x^3+\frac{1}{x^3}$

$2(4{\cos }^3\theta -3\cos \theta )=x^3+\frac{1}{x^3}$

$\therefore 2\cos 3\theta =x^3+\frac{1}{x^3}$