Binary Operations (Commutative and Associative). 
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. 
 A: 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.
