If $z^2 + z + 1 = 0$, then find the value of $\left(z + \frac{1}{z}\right)^2 + \left(z^2 + \frac{1}{z^2}\right)^2 + \left(z^3 + \frac{1}{z^3}\right)^2$.

The roots of $z^2 + z + 1 = 0$ are the non-real cube roots of unity, $\omega$ and $\omega^2$. Let $z = \omega$.
Note that $\frac{1}{\omega} = \omega^2$ and $\frac{1}{\omega^2} = \omega$, and $\omega^3 = 1$.

• Term 1: $\left(\omega + \frac{1}{\omega}\right)^2 = (\omega + \omega^2)^2 = (-1)^2 = 1$
• Term 2: $\left(\omega^2 + \frac{1}{\omega^2}\right)^2 = (\omega^2 + \omega)^2 = (-1)^2 = 1$
• Term 3: $\left(\omega^3 + \frac{1}{\omega^3}\right)^2 = (1 + 1)^2 = (2)^2 = 4$

Total Sum = $1 + 1 + 4 = 6$.

Answer: 6