Simplifying the difference of third roots

Question

How would I combine the two terms in the following expression? (ie. make it into a single term such as $f(x,a)^{1/3}$):

$(x+a)^{1/3} - (x-a)^{1/3}$ where $x,a$ are real numbers such that $a$, $x+a$ and $x-a$ are not zero

Answer 1

There is not any nice way to simplify this expression. In general, there is typically no way to simplify a sum of radicals.

In particular, there is no polynomial (or even differentiable function) $f(x,a)$ such that $$(x-a)^{1/3}+(x+a)^{1/3}=f(x,a)^{1/3}$$ for all $x,a\in\mathbb{R}$. There are many ways to prove this; here's one. Cubing both sides would give $$(x-a)+3(x-a)^{2/3}(x+a)^{1/3}+3(x-a)^{1/3}(x+a)^{2/3}+x+a=f(x,a)$$ or $$(x-a)^{2/3}(x+a)^{1/3}+(x-a)^{1/3}(x+a)^{2/3}=g(x,a)$$ where $g(x,a)=(f(x,a)-2x)/3$ is a polynomial. But this is impossible since if you fix $a\neq 0$ then the left-hand side is not differentiable with respect to $x$ at $a$.

Answer 2

I don't think you can get rid of the roots. For instance, if $x=1/2$ and $a=1/3$, then $$ (x+a)^{1/3} - (x-a)^{1/3}=\frac{5^{1/3}-1}{6^{1/3}}, $$ which is irrational. So whatever formula you can come up with, it will have to produce irrationals out of rationals, i.e. it will require roots.