Why the reals with the operation $x \bullet y = \sqrt[3]{x^3 + y^3}$ is a group? The operation above is a group for the real numbers, since 0 is the identity element, and the negative of any real number is its inverse, as it can be observed trivially. Associativity is less trivial, but it holds.
In fact, if we substitute 3 for any odd number (5, 7 ...), the operation satisfies the properties of the group. However, any even number fails.
Is there any geometric / analytic / ... interpretation why an operation like $x \bullet y = \sqrt[3]{x^3 + y^3}$ is associative and, as a consequence, it gives the structure of a group to the reals?
 A: If $f$ is any odd  bijection of the reals then the operation
$$x\cdot y=f^{-1}(f(x)+f(y))$$
makes the reals a group and $f$ an isomorphism from the additive group of the  reals to that group. In your case $f(x)=x^3$. Associativity follows from the fact that $f$ is a homomorphism. $0$ is the neutral element and $-x$  is the inverse of $x$. Here the fact that $f$ is odd is used.
A: For an arbitrary bijection $f\colon \mathbf R \to \mathbf R$, the operation $x*y = f^{-1}(f(x) + f(y))$ is a group law on $\mathbf R$. All this says is that if you rename each real number $x$ as $f(x)$ then you can convert the original group law $+$ into a group law $*$ so that $f$ is an isomorphism from $(\mathbf R, *)$ to $(\mathbf R,+)$. The intuition is algebraic, not geometric. There is nothing magical about $n$th roots for odd $n$ other than being a bijection.
The hyperbolic tangent function $\tanh \colon \mathbf R \to (-1,1)$ is a bijection that lets you transport addition on $\mathbf R$ to a group law on $(-1,1)$ that is used in special relativity (addition of velocities in one-dimensional motion). The inverse of this bijection, up to a scaling factor, is called “rapidity” in physics.
A: Short answer: because $\sqrt{x^2}\ne x$ for $x<0$.
Long answer, in which I prefer $\cdot$ to $\bullet$:
An operation satisfying $(x\cdot y)^n=x^n+y^n$ closes the reals, since if $n$ is odd we can take the $n$th root, & if $n$ is even we only try to take the $n$th root of something $\ge0$. And since$$((x\cdot y)\cdot z)^n=(x\cdot y)^n+z^n=x^n+y^n+z^n=(x\cdot(y\cdot z))^n,$$the operation associates. (Cancelling the power of $n$ is trivial since, even if $n$ is even, $\cdot$ is always defined to take the non-negative $n$th root anyway.) So at a minimum, we form a semigroup.
Since $x\cdot0=(x^n)^{1/n}$, for odd $n$ we also have $0$ as an identity, but for even $n$ we don't because $x\cdot0=|x|$, so it's not even a monoid, let alone a group. The last group axioms is inverses, which work for odd $n$ as you noted, but for even $n$ we have $x\cdot y\ge|x|$, so we don't have inverses either.
A: Let $G$ be any group, $X$ be any set, and $f: X \rightarrow G$ be any bijection. Then, we can transfer the group structure from $G$ to $X$ by setting $x \cdot y = f^{-1}(f(x) \cdot f(y))$. That is, we use the bijection $f$ to identify elements of $G$ and elements of $X$, and put a group structure on $X$ using this identification. I'll leave it as an exercise that this indeed defines a group structure on $X$.
Now, take $G=(\mathbb R,+)$, $X=\mathbb R$ and $f(x)=x^3$ to recover your case.
A: Hint:
Associativity simply results from the fact that both $\;(x\bullet y)\bullet z$ and  $\;x\bullet( y \bullet z)$ are equal to
$$\sqrt[\substack{\,\scriptstyle3\\}]{x^3 +y^3+z^3}.$$
