The sum of $$162^{th}$$ power of the roots of the equation $$x^3 - 2x^2 + 2x - 1 = 0$$ is ______.

We need to find the sum of the 162nd powers of the roots of $$x^3 - 2x^2 + 2x - 1 = 0$$.

We factor the polynomial: $$x^3 - 1 - 2x^2 + 2x = (x-1)(x^2+x+1) - 2x(x-1) = (x-1)(x^2+x+1-2x) = (x-1)(x^2-x+1) = 0$$.

The roots are $$x = 1$$ and the roots of $$x^2 - x + 1 = 0$$. Using the quadratic formula: $$x = \frac{1 \pm \sqrt{1-4}}{2} = \frac{1 \pm i\sqrt{3}}{2}$$.

These roots are $$e^{i\pi/3}$$ and $$e^{-i\pi/3}$$, which are primitive 6th roots of unity satisfying $$\omega^6 = 1$$.

Since $$162 = 6 \times 27$$, we have $$\omega^{162} = (\omega^6)^{27} = 1^{27} = 1$$. Similarly $$\bar{\omega}^{162} = 1$$.

Therefore the sum of the 162nd powers is $$1^{162} + \omega^{162} + \bar{\omega}^{162} = 1 + 1 + 1 = 3$$.