The equation whose roots are the $$n^{th}$$ power of the roots of the equation $${x^2} - 2x\cos \theta + 1 = 0$$     is given by

Solution :
The roots of the given equation are
$$x = \frac{{2\cos \theta \pm \sqrt {4{{\cos }^2}\theta - 4} }}{2} = \cos \theta \pm i\sin \theta $$
Let, $$\alpha = \cos \theta + i\sin \theta \,\& \,\beta = \cos \theta - i\sin \theta $$
Then, $${\alpha ^n} = \cos n\theta + i\sin n\theta $$
$${\beta ^n} = \cos n\theta - i\sin n\theta $$
[Using De Moivre Theorem]
$${\alpha ^n} + {\beta ^n} = 2\cos n\theta \,\,{\text{and }}{\alpha ^n} \cdot {\beta ^n} = 1$$
∴ The required equation is
$${x^2} - 2x\cos n\theta + 1 = 0$$