Stationary points of $(x+y+z)^3 - 3(x+y+z)-24xyz +a^3$

Question

The solution in the book says that there are 2 stationary points $(1,1,1)$ and $(-1,-1,-1) $.

Why $(0,1,0)$, $(0,-1,0)$,$(1,0,0)$ and $(-1,0,0)$ are not the stationary points ?

Answer 1

You must solve the system $$3(x+y+z)^2-3-24yz=0$$ $$3(x+y+z)^2-3-24xz=0$$ $$3(x+y+z)^2-3-24yx=0$$