If $x^3 - 2x^2 + 2x - 1 = 0$ has roots $(\alpha, \beta, \gamma)$, then find $(\alpha^{162} + \beta^{162} + \gamma^{162})$.

**Solution:** Given cubic: $x^3 - 2x^2 + 2x - 1 = 0$. Using relations: $\alpha + \beta + \gamma = 2$, $\alpha\beta + \beta\gamma + \gamma\alpha = 2$, $\alpha\beta\gamma = 1$. Form recurrence: $a_n = \alpha^n + \beta^n + \gamma^n$. Then $a_0 = 3,\ a_1 = 2$ and by the cubic relation: $a_n = 2a_{n-1} - 2a_{n-2} + a_{n-3}.$ We get periodic pattern with period 3, thus $a_{162} = a_0 = 3.$ $\boxed{\text{Answer: (C) 3}}$