Simplify this expansion : $\omega=(28+(\frac{5290}{3})^{\frac{3}{2}})^{\frac{1}{3}}+(28-(\frac{5290}{3})^{\frac{3}{2}})^{\frac{1}{3}}$ Find a simple closed form of : 
$\omega=(28+(\frac{5290}{3})^{\frac{3}{2}})^{\frac{1}{3}}+(28-(\frac{5290}{3})^{\frac{3}{2}})^{\frac{1}{3}}$
My try : 
Let : 
$A=(28+(\frac{5290}{3})^{\frac{3}{2}})^{\frac{1}{3}}$ 
And 
$B=(28-(\frac{5290}{3})^{\frac{3}{2}})^{\frac{1}{3}}$ 
Now : 
$A^{3}+B^{3}=56$ 
But how I can now find $A$ and $B$ ? 
 A: I would write
$$(28+x)^{1/3}+(28-x)^{1/3}=s$$ and now raise this to the power three.
A: Given
$$\omega=\sqrt[3]{28+(\frac{5290}{3})^{\frac{3}{2}}}+\sqrt[3]{28-(\frac{5290}{3})^{\frac{3}{2}}}$$
Let $A=\sqrt[3]{28+(\frac{5290}{3})^{\frac{3}{2}}}\Rightarrow A^3=28+(\frac{5290}{3})^{\frac{3}{2}}$
and $B=\sqrt[3]{28-(\frac{5290}{3})^{\frac{3}{2}}}\Rightarrow B^3=28-(\frac{5290}{3})^{\frac{3}{2}}$.
Notice that $A^3+B^3=56$.
Also, $AB=\sqrt[3]{\big(28+(\frac{5290}{3})^{\frac{3}{2}}\big)\big(28-(\frac{5290}{3})^{\frac{3}{2}}\big)}=\sqrt[3]{28^2-(\frac{5290}{3})^3}=-\frac23\sqrt[3]{18504483479}$.
So we can write
\begin{align}
\omega&=A+B\\
\omega^3&=A^3+3A^2B+3AB^2+B^3=A^3+B^3+3AB\cdot(A+B)\\
\omega^3&=56+3(-\frac23\sqrt[3]{18504483479})\ \omega\\
\omega^3&=56-2\sqrt[3]{18504483479}\ \omega\\
\end{align}
At the end of the day, you are solving the cubic equation
$$\omega^3+2\sqrt[3]{18504483479}\ \omega-56=0\\$$
There has got to be a typo somewhere.
A: As you did, let
$$ A=(28+(\frac{5290}{3})^{\frac{3}{2}})^{\frac{1}{3}}, B=(28-(\frac{5290}{3})^{\frac{3}{2}})^{\frac{1}{3}}. $$
Then
$$ A^3+B^3=56, AB=-\frac23 (18504483479)^{1/3}. $$
Let $z=A+B$ and then
$$ A^3+B^3=(A+B)[(A+B)^2-3AB]=z^3+2(18504483479)^{1/3}z=56. $$
So $z=A+B$ satisfies
$$ z^3+2(18504483479)^{1/3}z-56 = 0. $$
