Show that $\alpha^3 = 10\alpha - 24$ given that $\alpha + \beta = 4$ and $\alpha \beta = 6$ How to show that $\alpha^3 = 10\alpha - 24$ given that $\alpha + \beta = 4$ and $\alpha \beta = 6$
As given in the title. I tried shifting the RHS to the left but I'm not sure if factorization is the way to go. There are earlier parts to this question but I'm not sure if they will be helpful so I'm excluding them.
Thank you.
(Question is under sum and product of roots for a quadratic equation.)

Sum of roots = $ \alpha + \beta + \gamma =$
$ \frac{-b}{a} = \frac{-0}{1} = 0  $
But, $ \alpha + \beta = 4 $ $\implies \gamma = -4 $
Product of roots $\alpha * \beta * \gamma = $
$ \frac{-d}{a} = -24 $ 
Sum of roots taken two at a time = $ \alpha\beta + \beta\alpha + \gamma\alpha =$
$ \frac{c}{a} = \frac{10}{1} = 10 $
Now, for a cubic equation:
$$ x^3 - (\alpha+\beta+\gamma)x^2+(\alpha\beta + \beta\alpha + \gamma\alpha )x - (\alpha\beta\gamma)=0 $$
Or 
$$ \alpha^3-10\alpha+24=0 $$
