How to find the critical points of $f(x,y,z) = x^4 + y^4 + z^4 - 4xyz$? So I'm trying to find the critical points of this function $f(x,y,z) = x^4 + y^4 + z^4 - 4xyz$, to do that I try to find the points where the gradient of f is equal to $(0,0)$, though I can't solve the systems of equations, is this even the right way to do it?
Here's the system of equations that I'm trying to solve:
$4x^3-4yz = 0$
$4y^3-4xz = 0$
$4z^3-4xy = 0$
I first do this:
$x^3 = yz$
$y^3 = xz$
$z^3 = xy$
Then this:
$y = \dfrac{x^3}{z}$
$z = \dfrac{y^3}{x}$
$x = \dfrac{z^3}{y}$
But then I just go in circles by replacing each variable, I'm probably doing something wrong but I can't see it...
 A: Hint:
$$x^9=y^3z^3=x^2yz=x^5\implies x=0\quad\text{or}\quad x^4=1.$$

 This results in 17 solutions: $(0,0,0)$ and 16 non-zero solutions, which are all possible permutations of $(1,1,1),(1,-1,-1),(1,i,-i),(-1,i,i),(-1,-i,-i).$

A: Once I have
$x^3=yz$
$y^3=xz$ and
$z^3=xy$
I would notice that dividing the first equation by the second eliminates z: $\frac{x^3}{y^3}= \frac{y}{x}$ so that $x^4= y^4$.  That has the two solution y= x and y= -x.
If y= x, the three equations are
$x^3= xz$ so $x^2= z$
$x^3= xz$ again, and
$z^3= x^2$.
Since $z= x^2$, $x^6= x^2$.
$x^6- x^2= x^2(x^4- 1)= x^2(x^2- 1)(x^2+ 1)= 0.
there are three solutions.
1)x= 0, y= 0, z= 0.
2)x= 1, y= 1, z= 1.
3)x= -1, y= -1, z= -1.
Now, if y= -x, then the three equations become
$x^3= -xz$ so that $x^2= -z$
$-x^3= xz$ again and
$z^3= -x^2$.
Since $z= -x^2$, $z^3= -x^6= -x^2$
$x^6- x^2= x^2(x^4- 1)= x^2(x^2- 1)(x^2+ 1)= 0$
That gives the same x values as before but now y= -x and $z= -x^2$.  We have

*

*x= 0, y= 0, z= 0

*x= 1, y= -1, z= -1

*x= -1, y= 1, z= -1.

A: Hint: Multiply the first two equations with $x$ respectively $y$, write $x^4=y^4=xyz$ and solve $x^4=y^4$ which will rezult in $x=\pm y$, and so on.
A: Hint:
$$x^3 = yz \iff x^4 = xyz$$
$$y^3 = xz \iff y^4 = xyz$$
$$z^3 = xy \iff z^4 = xyz$$
Can you finish?
