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 $\alpha^2 + \beta^2 + \gamma^2$.

$ x(3x^2-4)=8$
Squaring we get $x^2(3x^2-4)^2=8^2$
Let $x^2=y\implies y(3y-4)^2=64\iff9y^3-24y^2+16y-64=0$
whose roots are $a^2,b^2,c^2$
$\implies a^2+b^2+c^2=\dfrac{24}9=?$
