Simplify this expansion : $\omega=(28+(\frac{5290}{3})^{\frac{3}{2}})^{\frac{1}{3}}+(28-(\frac{5290}{3})^{\frac{3}{2}})^{\frac{1}{3}}$

Question

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 $B$ ?

Answer 1

I would write $$(28+x)^{1/3}+(28-x)^{1/3}=s$$ and now raise this to the power three.

Answer 2

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.

Answer 3

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