Critical points of the function $\omega = (x^3 +1)(y^3+1)$

Question

I have a question stating

Find and classify all the critical points of $\omega = (x^3 +1)(y^3+1)$

First I expanded it to the form $\omega = x^3y^3 +x^3 +y^3 +1$ to make it easier for me. I then found the derivatives: $$\frac{\partial\omega}{\partial x} = 3y^3x^2 +3x^2 \\ \frac{\partial\omega}{\partial y} = 3x^3y^2 +3y^3 $$

I know that the critical points are where both partial derivatives equal $0$, so I must solve the simultaneous equations: $$\begin{cases} 3y^3x^2+3x^2=0 \\ 3y^2x^3+3y^2=0 \end{cases}$$

I don't know how to solve these simultaneous equations. I tried factorising to get, $$\begin{cases} 3x^2(y^3 +1) = 0 \\ 3y^2(x^3 +1) = 0 \end{cases}$$ but I don't see where to go from here. Could anyone point me in the right direction?

Answer 1

From $$\begin{cases} 3x^2(y^3 +1) = 0 \\ 3y^2(x^3 +1) = 0 \end{cases}$$

You solve the first equation and get $x=0$ and $y=-1$.

Similarly you solve the second equation and get $y=0$ and $x=-1$.

That gives you the critical points of $$(0,0),(-1,-1).$$

Now at each point you need to classify the critical point into local max, local min, and saddle points.

Answer 2

Hint

Substituting the solution $x=0$ from the first equation in the second we find $y=0$ so one solution is $(0,0)$.

And substituting $y=-1$ from the first to the second we find another solution.....