Way to go, running-gag! Although I don't think that was quite what NowisForever had in mind.

NowisForever, here's the hard way:

\(\displaystyle (a+ b)^3= a^3+ 3a^2b+ 3ab^2+ b^3\) so, in particular, \(\displaystyle x^3= \left(\sqrt[3]{10+ \sqrt{108}}+ \sqrt[3]{10- \sqrt{108}}\right)^3\)\(\displaystyle = 10+ \sqrt{108}+ 3\sqrt[3]{10+\sqrt{108})^2(10- \sqrt{108}}\)\(\displaystyle + 3\sqrt[3]{(10+ \sqrt{108})(10- \sqrt{108})^2+ 10- \sqrt{108}}\)

Now, \(\displaystyle (10+ \sqrt{108})^2(10- \sqrt{108})= (10+ \sqrt{108})(10+ \sqrt{108})(10- \sqrt{108})\)\(\displaystyle = (10+ \sqrt{108})(100- 108)= -8(10+ \sqrt{108}\) and, equivalently, \(\displaystyle (10+ \sqrt{108})(10- \sqrt{108})^2= -8(10-\sqrt{108}\).

So that cube becomes \(\displaystyle x^3= 20- 6\sqrt[3]{10+ \sqrt{108}}- 6\sqrt[3]{10- \sqrt{108}}\).

But \(\displaystyle 6x= 6\sqrt[3]{10+ \sqrt{108}}+ 6\sqrt[3]{10- \sqrt{108}}\) so that \(\displaystyle x^3+ 6x= 20\) and the equation follows.