Consider the law of composition $(x,y) \rightarrow \sqrt[3]{x^3+y^3}$ on the set of real numbers. Show that this law of composition defines a group isomorphic to the additive group ($G$) of $\mathbb{R}$.
My attempt: So I know the additive group on $\mathbb{R}$ is just the set of all real numbers that 1) have an identity such that : $x+e = e+x$ for all $x \in G$ i.e. ($e=0$) 2) have an additive inverse 3) are associative
My confusion arises because Im not sure what composition of two elements for instance $(x_1,y_1) \circ (x_2,y_2)$ looks like. Do I add $\sqrt[3]{x_1^3+y_1^3} + \sqrt[3]{x_2^3+y_2^3}$ which then has two values both in $\mathbb{R}$ and then I can see that we are just adding real numbers, which makes sense for the isomorphism, or is the law of composition $\sqrt[3]{(x_1+x_2)^3+(y_1+y_2)^3}$, in which case im a bit lost.
How do I know which is the proper way to compose in the original group? Im heavily leaning towards the first option, because then we can have that $(0,0)$ be our identity element, and $-(x,y)$ be our inverse and we have a subgroup, but I dont understand why this is the law of composition to choose.
In terms of showing that it is an isomorphism, unfortunatley, I am completely lost.
Any help is appreciated, thanks!