Question

If ${(x+\frac{1}{x})}^2=3$ then find the value of $x^6$.

Answer 1

Given,

${(x+\frac{1}{x})}^2=3$

or, $x^2+2+\frac{1}{x^2}=3$

or, $x^4-x^2+1=0$

or, $x^2=\frac{1\pm i\sqrt{3}}{2}$ [ Where $i^2=-1$]

or, $x^2=\frac{1}{2}\pm i\frac{\sqrt{3}}{2}$

or, $x^2=\cos \frac{\pi }{3}\pm i\sin \frac{\pi }{3}$.

Now cubing both sides we get,

$x^6=\cos \pi \pm i\sin \pi$ [ Using De Movire's theorem]

or, $x^6=-1$.