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 $B$ ? 

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
$$ z^3+2(18504483479)^{1/3}z-56 = 0. $$
