Fast way to solve $4 = \sqrt[3] {x+10}-\sqrt[3] {x-10}$

If $\sqrt[3]{x-10}=t$, then the equation is

$$4+t=\sqrt[3]{t^3+20}.$$

So $$64+48t+12t^2+t^3=t^3+20,$$ $$\iff 12t^2+48t+44=0,$$ which gives $ t=(-24\pm \sqrt{48})/12=-2\pm \sqrt{3}/3$ and $$x=t^3+10=\pm 7.12065332001...$$

Use this formula $$\boxed{(a-b)^3 = a^3-3ab(a-b)-b^3}$$

$$4 = \sqrt[3] {x+10}-\sqrt[3] {x-10}\;\;\;/^3$$

$$64 = x+10 -3\sqrt [3]{(x+10)(x-10)}(\underbrace{\sqrt[3] {x+10}-\sqrt[3] {x-10}}_{4})-x+10$$

So $$11 = -3\sqrt [3]{x^2-100}\implies x = \pm\sqrt{{-11^3\over 27}+100}$$