Let $*$ be a binary operation on $\mathbb R$ given by $x*y = (x^{1/3}+ y^{1/3})^3$.
Determine if it is commutative and associative.
I know how to prove it is commutative.
\begin{align*}x * y &= (x^{1/3} + y^{1/3})^3\\ &= (y^{1/3} + x^{1/3})^3 = y * x.\end{align*} Therefore, it is commutative.
But how do I prove if it is associative? I know that $(x * y) * z$ should be equal to $x * (y * z)$. But I don't know how to start the proof.
You really just have to write out the two expressions $(x*y)*z$ and $x*(y*z)$. If you do, you get: \begin{align*}(x*y)*z&=((x*y)^{1/3}+z^{1/3})^{3}\\ &=(\left((x^{1/3}+y^{1/3})^3\right)^{1/3} +z^{1/3})^3\\ &=(x^{1/3}+y^{1/3}+z^{1/3})^3.\end{align*} You can write out the other expression, and see that the same cancellation takes place and that they are equal.
There is a more general fact at play here however: The map $f:\mathbb R\rightarrow \mathbb R$ given by $f(x)=x^{1/3}$ is a bijection. If you already know that addition is commutative and associative, you can show the same of this operation if you note that $$f(x*y)=f(x)+f(y)$$ Since then you can write $$f(x*y)=f(x)+f(y)=f(y)+f(x)=f(y*x)$$ and then, since $f$ is a bijection, that means $x*y=y*x$. Similarly, for associativity, you can just write $$f((x*y)*z)=f(x*y)+f(z)=f(x)+f(y)+f(z)=f(x)+f(y*z)=f(x*(y*z))$$ therefore $(x*y)*z=x*(y*z)$.
This is a nice way to view this sort of question, since it tells you that $*$ basically has the exact same properties as the usual addition on $\mathbb R$. More formally, these operations are isomorphic because they satisfy this relation, and this means they are essentially the same, except that we "relabelled" the points in $\mathbb R$ somehow.