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 ?
 A: Let $\alpha =\dfrac {56}2, \beta=\dfrac 1 2\sqrt{\dfrac{84640}{27}}$ and $\gamma =\root 3\of {\alpha -\beta}+\root 3\of {\alpha +\beta}$. 
All the numbers taken are supposed to be real.
Now note that 
$$\begin{align} 
\gamma ^3&=\left(\root 3\of {\alpha -\beta}\right)^3+3\root 3\of {\alpha -\beta}^2\root 3\of {\alpha +\beta}+3\root 3\of {\alpha -\beta}\root 3\of {\alpha +\beta}^2+\left(\root 3\of {\alpha +\beta}\right)^3\\
&=\alpha -\beta +3\root 3\of {\alpha -\beta}\root 3\of {\alpha +\beta}\Bigl(\underbrace{\root 3\of {\alpha -\beta}+\root 3\of {\alpha +\beta}}_{\huge =\gamma}\Bigr)+\alpha +\beta\\
&=2\alpha+3\root 3\of {\alpha ^2-\beta ^2}\gamma.
\end{align}$$
Since $\alpha ^2=784$ and $\beta ^2=\dfrac{21160}{27}$ it follows that $\alpha ^2-\beta ^2=\dfrac{21168}{27}-\dfrac{21160}{27}=\dfrac 8{27}$. 
Thus $\root 3\of {\alpha ^2+\beta ^2}=\dfrac 2 3$.
One concludes that $\gamma$ is such that $\gamma^3=56+2\gamma$, hence $\gamma$ is a (the only) real root of $x^3-2x-56$.
A: 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 $4$. 
A: cubing both sides, 
$56+(\frac{1}{4}(56^2-\frac{84640}{27})
)^{1/3}(4)=64$ simplify and you will get the result.
