How to solve the equation $x^3+y^3=0$ for real numbers $x$ and $y$?

Question

I'm finding stationary points of the function $f(x,y)=2(x-y)^2-x^4-y^4$, but stuck in the equation $x^3+y^3=0$ while solving the equations $f_x=0$ and $f_y=0$.

Please help me. Thanks in advance.

Answer 1

HINT: Notice, $$x^3+y^3=(x+y)(x^2+y^2-xy)$$ Then, we have $$(x+y)(x^2+y^2-xy)=0$$

Answer 2

Starting with $x^3+y^3=0$, we have $$x^3 = -y^3 =(-y)^3$$ so $$x=-y$$ since cubing is a one-to-one function. Indeed, the line $x=-y$ is the solution set to the equation $x^3+y^3=0$