We’d like to get rid of the cube roots in the equation so that we can simplify. Cubing both sides might help, but when we expand the lhs we’ll get ‘cross terms’ (the $3ab(a+b)$ in the expansion of $(a+b)^3$). At this point in solving the problem, I have an inkling that the cross terms might simplify to something nice, but I’m not sure; it’s good to go ahead with an idea and see what happens even if you don’t really have a full plan. Let’s try it.

Let $a = \sqrt[3]{x + \sqrt{x^2-27}}$, $b=\sqrt[3]{x - \sqrt{x^2 - 27}}$. The original equation rewritten in terms of $a$ and $b$ is $a+b=5$. If we cube both sides, we get: \begin{align*} (a+b)^3 &= 125\\ a^3 + b^3 + 3ab(a+b) &= 125 \end{align*} Now let’s simplify the $a^3+b^3$, $ab$, and $a+b$: \begin{align*} a^3 + b^3 &= x + \sqrt{x^2-27} + x - \sqrt{x^2 - 27}\\ &= 2x \end{align*} \begin{align*} ab &= \sqrt[3]{x + \sqrt{x^2-27}}\times\sqrt[3]{x - \sqrt{x^2 - 27}}\\ &= \sqrt[3]{(x + \sqrt{x^2-27})(x - \sqrt{x^2 - 27})}\\ &= \sqrt[3]{x^2 - (x^2 - 27)}\quad\text{(using difference of two squares expansion)}\\ &= \sqrt[3]{27}\\ &= 3 \end{align*} And of course $a+b=5$ from the original equation. Putting all this together gives: \begin{align*} a^3 + b^3 + 3ab(a+b) &= 125\\ 2x + 3\times3\times5 &= 125\\ 2x &= 80\\ x &= 40 \end{align*}