Proving an expression is $4$ 
Is $\displaystyle\sqrt[\huge3]{\frac{1}{2} \left(56-\sqrt{\frac{84640}{27}}\right)}+\sqrt[\huge 3]{\frac{1}{2} \left(\sqrt{\frac{84640}{27}}+56\right)}=4$ true ?

This was asked during an oral examination where calc and CAS are forbidden.
Mathematica seems to say it's true.
Can someone find a nice proof ?

What are these guys sadists ? How is one supposed to answer this in an oral exam ?
Anyway here is another solution, we recognise the expression as the form of Cardano's solution to the cubic $x^3+px+q=0$
$$\frac{1}{3}\left( \sqrt[3]{-\frac{27}{2}q +
 \frac{3\sqrt{-3D}}{2}}+
 \sqrt[3]{-\frac{27}{2}q - 
\frac{3\sqrt{-3D}}{2}} \right)$$ where $D=-27q^2-4p^3$
if we multiply the $\frac{1}{3}$ through we have
$$ \sqrt[3]{\frac{1}{2}(-q +
 \sqrt{ \frac{-3D}{27} })}+
 \sqrt[3]{\frac{1}{2}(q - \sqrt{\frac{-D}{27}}} )$$
So we recognise immediately that $q=-56$ and $-D=84640$ so 
$$27q^2+4p^3=84640$$ and this gives $p=-2$
So the cubic in question is $x^3-2x-56$ which has the root $4$. 
