$x^3+ax=b\implies x=\sqrt[3]{\frac b2+\sqrt{(\frac b2)^2+(\frac a3)^3}}+\sqrt[3]{\frac b2-\sqrt{\left(\frac b2\right)^2+\left(\frac a3\right)^3}}$

Question

I was told that

$x=\sqrt[3]{\frac{b}{2}+\sqrt{\left(\frac{b}{2}\right)^{2}+\left(\frac{a}{3}\right)^{3}}}+\sqrt[3]{\frac{b}{2}-\sqrt{\left(\frac{b}{2}\right)^{2}+\left(\frac{a}{3}\right)^{3}}}$

solves the equation $x^{3}+ax=b$

and there is a proof

I am trying to find it but i am stuck

all i have been able to do is:

$\left(\sqrt[3]{\frac{b}{2}+\sqrt{\frac{b}{2^{2}}^{2}+\frac{a}{3^{3}}^{3}}}+\sqrt[3]{\frac{b}{2}-\sqrt{\frac{b}{2^{2}}^{2}+\frac{a}{3^{3}}^{3}}}\right)^{3}+a\cdot\sqrt[3]{\frac{b}{2}+\sqrt{\frac{b}{2^{2}}^{2}+\frac{a}{3^{3}}^{3}}}+a\cdot\sqrt[3]{\frac{b}{2}-\sqrt{\frac{b}{2^{2}}^{2}+\frac{a}{3^{3}}^{3}}}=b$