A specific question for three sets to be equal (Calculation Problem) Let $x$, $y$, $z$ be three different numbers, and $A_1=\{x,\ y,\ z\}$, $A_2=\{x^2,\ y^2,\ z^2\}$, $A_3=\{xy,\ yz,\ zx\}$. If $A_1=A_2=A_3$, what is the set?

The answer seems to be $\{1,\ w,\ w^2\}$, where $w=\frac{-1\pm\sqrt{3}i}{2}$, but I have no idea about how to get this result.

Thanks.
 A: At least one of the elements $\{x,y,z\}$ is non-zero. 
Assume that $x \neq 0.$
Then, since $x^2 \in \{xy,yz,zx\}$, and $x,y,z$ are all distinct, you must have that $x^2 = yz \implies 0 \neq y,z$.
Then, similar analysis gives:

*

*$x^2 = yz$

*$y^2 = xz$

*$z^2 = xy$.

This implies that $xyz = x^3 = y^3 = z^3.$
Letting $r_1,r_2,r_3$ denote the $3$ roots of $z^3 - 1= 0,$ you can immediately conclude that
$y = (r_j \times x), z = (r_k \times x)$, where $j,k \in \{1,2,3\}$ and $j \neq k.$
However, this implies that $|x| = |y| = |z|.$
Therefore, since $x^2 \in \{x,y,z\}$, you can conclude that $|x| = |x^2| = |x|^2 \implies |x| = 1.$
So, at this point, you know that $x,y,z$ are all elements of the unit circle, and that the absolute value of the difference in their arguments is $\dfrac{2\pi}{3}$.
It only remains to show that $x,y,z$ must specifically be $\{r_1,r_2,r_3\}.$
It is sufficient to show that (for example) $x$ must be an element in $\{r_1,r_2,r_3\}$.
You know that $x^2 \in \{x,y,z\}$.
This implies that $x^2$ must be expressible as $xr_k : k \in \{1,2,3\}.$  This implies that $x = r_k$, which completes the analysis.
