Find the roots of $3x^3-4x-8$ It is given that $\alpha$, $\beta$ and $\gamma$ are the roots of the polynomial $3x^3-4x-8$.
I have been asked to calculate the value of $\alpha^2 + \beta^2 + \gamma^2$.
However I am unsure how to find these roots, seeing as though I haven't been given a root to start with. 
I began by identifying that  
$\alpha + \beta + \gamma = 0$
$\alpha\beta + \alpha\gamma + \beta\gamma = -4/3$
$\alpha\beta\gamma = 8/3$
However I am unsure how to continue to find $\alpha^2 + \beta^2 + \gamma^2$.

Yet another approach: $\,\dfrac{1}{\alpha}, \dfrac{1}{\beta}, \dfrac{1}{\gamma}\,$ are the roots of $\,8x^3+4x^2-3\,$, so $\,\dfrac{1}{\alpha}+ \dfrac{1}{\beta}+ \dfrac{1}{\gamma}=-\dfrac{1}{2}\,$.
But $\,3\alpha^3=4\alpha+8 \iff \alpha^2=\dfrac{4}{3} + \dfrac{8}{3\alpha}\,$, and therefore: $$\alpha^2+\beta^2+\gamma^2=3 \cdot \frac{4}{3}+ \frac{8}{3}\left(\frac{1}{\alpha}+ \frac{1}{\beta}+ \frac{1}{\gamma}\right)=4-\frac{4}{3} = \frac{8}{3}$$
