Functions over Relations
why differentiate between a function and a relation?
Please note that mathematical objects are usually defined because their existence is a need to facilitate dealing with some math subjects. When you see that a math concept is often appeared in mathematical literature, you can conclude that it has a significant role and many applications inside and outside of (pure) mathematics.
The concept "relation" has been defined to show mathematical relationships between mathematical objects. A "function" is a special kind of a relation that for each input there is only one output; in fact, functions are well-behaved relations because we can control the outputs of a function by controlling its inputs, which is a very important fact in developing calculus subjects such as limit, differentiation, integration, and so on. Looking at various subjects inside and outside of (pure) mathematics, we can find that almost all relations are functions or can be written as a union of some functions.
For example, consider the circle $C:y=\pm \sqrt{1-x^2}$. This relation is not a function because for each input there are two outputs. However, we can write it as the union of the following functions:$$y=\begin{cases}f_1(x)=\sqrt{1-x^2} \\ f_2(x)=-\sqrt{1-x^2} \end{cases} \quad \Rightarrow \quad C= f_1 \cup f_2.$$Now, we can apply any facts about functions to the function pieces of a relation. Please note that almost all (applied) facts about functions are local properties, so we can use them to treat a relation as a function.
Please note that there is a general principle:
Generality of a concept is inversely related to the information we know about the concept.
This principle not only holds in mathematics but also in other branches of knowledge.
Functions are less general than relations, but we have much more information about functions than relations. Almost all facts in many various (applied) mathematics subjects are expressed in terms of functions, and most of them cannot be expressed in terms of relations, and even if they can, they become awkward; also as mentioned above, many relations can be written as a union of functions.
Complexity of Inverting
Why does combining them seems to make inversion that much harder?
I think the question is a special case of the following question:
Why are there many mathematics problems which most people can understand easily but have very difficult (or do not have) solutions?
The answer is, because we have a few known facts (Mathematicians call them "axioms") and we have to prove any results from them.
Let us change your mentioned example. The inverse of the functions $g(x)=x^5$ and $h(x)=-x$ can be easily found. So, why can we not find the inverse of the following function easily:$$f(x)=g(x)+h(x)=x^5-x$$(I only added two "elementary functions")?
Please note that there are some facts written in less than ten words but their proofs have hundred pages; there are also some facts which most people can understand but a few mathematicians can understand their proofs.
Mathematics is an axiomatic theory. It does not ask people to find easy proofs for its facts; it only wants them to prove results from only a few axioms.
Could you please provide a link to the actual talk you saw so others can check you heard things correctly? I ask this because the nonexistence of a rational-sided right triangle with area 1 is well-known in number theory to be equivalent to the equation $x^4 + y^4 = z^2$ having no solution in nonzero (equivalently, positive) integers $x$, $y$, and $z$. That equation is close to, but not the same as, the Fermat equation of degree $4$ (which defines a projective curve of genus 3). I suspect that what Wiles had in mind is the connection between rational-sided right triangles of area $1$ and FLT for $n = 4$, not $n=3$.
FLT for $n=3$ is equivalent to the equation $y^2 = x^3 + 16$ having no solution in nonzero rational numbers, as shown in the last paragraph here. The nonexistence of a rational-sided right triangle of area $1$ is equivalent to the equation $y^2 = x^3 - x$ having no solution in nonzero rational numbers $x$ and $y$. The (elliptic) curves $y^2 = x^3 - x$ and $y^2 = x^3 + 16$ are not the same thing (and are not isogenous).
Update: In Weil's "Number Theory: An Approach Through History from Hammurapi to Legendre," Weil discusses the Fermat equation of degree 3 on pp. 114-117, saying Fermat claimed to prove it by descent but never revealed what he did, and then Weil discusses Euler's work on this equation. Euler is generally credited with the first solution of FLT for degree $3$. (His approach had a gap involving subtleties about $\mathbf Z[\sqrt{-3}]$, which was later filled in.) On pp. 76-79 Weil discusses letters of Fermat and Frenicle about there not being a rational-sided right triangle with area 1 or 2 (equivalently, area equal to a square or twice a square) and he points out the connection of these to nonsolvability of $x^4 \pm y^4 = z^2$ in positive integers. In algebraic number theory books, the Fermat equation of degree 4 is always discussed in connection with $x^4 + y^4 = z^2$. That equation is never brought up when discussing the Fermat equation of degree $3$. Fermat is credited with showing there is no rational-sided right triangle with area 1 and with solving FLT in degree 4 because the latter follows from the former, but he is not credited with solving FLT in degree $3$ despite claiming he could do so since we have no details of such a proof by him. In short, I think Wiles made a mistake by linking right triangles with area $1$ to FLT in degree $3$.
Best Answer
Hint: Take the sixth power of both sides.
Added: (for completeness, after the OP had used the hint to solve the problem) Since $\sqrt{a^2+b^2}$ is by definition non-negative, we have $$\sqrt{a^2+b^2}=\sqrt[3]{a^3+b^3} \qquad\text{iff}\qquad \left(\sqrt{a^2+b^2}\right)^6=\left(\sqrt[3]{a^3+b^3}\right)^6.$$ Equivalently, we want to find the solutions of $$(a^2+b^2)^3=(a^3+b^3)^2.$$ in non-zero reals. Expand. We get $$a^6 +3a^4b^2+3a^2b^4+b^6=a^6+2a^3b^3+b^6,$$ which is equivalent to $$3a^4b^2+3a^2b^4=2a^3b^3.$$ Since we are looking for solutions where neither $a$ nor $b$ is equal to $0$, we are looking for non-zero real solutions of $$3a^2+3b^2=2ab.$$ This equation has no non-zero real solutions. For by completing the square we get $$a^2+b^2-\frac{2ab}{3}=\left(a-\frac{b}{3}\right)^2+\frac{8b^2}{9}.$$ In order for the right hand side to be $0$, both terms must be $0$. In particular, $b$ must be $0$, and therefore so must $a$.
Comment: You may find the following related idea interesting. Let $p>1$, and let $t$ be positive. We will prove that $$1+t> (1+t^p)^{1/p}. \qquad\qquad(\ast)$$ Let $f(t)=1+t -(1+t^p)^{1/p}$. Note that $f(0)=0$. So it is enough to show that for positive $t$, $f(t)$ is an increasing function. To do this, we use the derivative: $$f'(t)=1 -\frac{t^{p-1}}{(1+t^p)^{(p-1)/p}}.$$ For positive $t$, $f'(t)$ is positive, since $(1+t^p)^{(p-1)/p}>(t^p)^{(p-1)/p}=t^{p-1}$. This completes the proof of $(\ast)$.
To apply $(\ast)$ to our problem, observe first that a solution of our equation with positive $a$ yields a solution of $(1+(b/a)^2)^{1/2}=(1+(b/a)^3)^{1/3}$. It is easy to check that $b/a$ cannot be negative. Now let $t=(b/a)^2$. We obtain the equation $1+t=(1+t^{3/2})^{2/3}$, which, by $(\ast)$, cannot hold for positive $t$.
We reduced the problem to one variable in order to use familiar tools. But there are important generalizations to several variables.