You're on a roll — To prove that $\alpha^3 + \beta^3 = -\frac{8}{27}$, we use the values of $\alpha + \beta$ and $\alpha\beta$ derived from the original quadratic equation $3x^2 - 4x + 2 = 0$. Step 1: Recall the sum and product of roots. For the equation $3x^2 - 4x + 2 = 0$, the roots are $\alpha$ and $\beta$. $$ \alpha + \beta = -\frac{-4}{3} = \frac{4}{3} $$ $$ \alpha\beta = \frac{2}{3} $$ Step 2: Use the identity for the sum of cubes. The correct identity for $\alpha^3 + \beta^3$ is: $$ \alpha^3 + \beta^3 = (\alpha + \beta)((\alpha + \beta)^2 - 3\alpha\beta) $$ Step 3: Substitute the values of $\alpha + \beta$ and $\alpha\beta$ into the identity. $$ \alpha^3 + \beta^3 = \left(\frac{4}{3}\right)\left(\left(\frac{4}{3}\right)^2 - 3\left(\frac{2}{3}\right)\right) $$ $$ \alpha^3 + \beta^3 = \left(\frac{4}{3}\right)\left(\frac{16}{9} - \frac{6}{3}\right) $$ $$ \alpha^3 + \beta^3 = \left(\frac{4}{3}\right)\left(\frac{16}{9} - \frac{18}{9}\right) $$ $$ \alpha^3 + \beta^3 = \left(\frac{4}{3}\right)\left(-\frac{2}{9}\right) $$ $$ \alpha^3 + \beta^3 = -\frac{8}{27} $$ Thus, it is proven that $\boxed{\alpha^3 + \beta^3 = -\frac{8}{27}}$.