I've been trying to solve the following problem:
Given the curve $F_{\mu} =X^3+Y^3+Z^3+ 3\mu XYZ$ in $\Bbb{P}^{2}_{\Bbb{C}}$, show that $F_{\mu}$ has singular points only if and only if $\mu^3=-1$.
I tried to use the following: $F_{\mu}$ has singular points if and only if $F_X=F_Y=F_Z=0$. And hence:
$$3 X^2+3\mu Y Z=0\\ 3Y^2+ 3\mu X Z=0\\3 Z^2+ 3\mu X Y=0$$
Question 1: I think we can prove that if $\mu^3=-1$, then we can find solutions, by trying all possible values and hence, there are singular values. Does that make sense? It seems it could work but I've been trying with no success.
Question 2: What do I do in the other direction?
Your curve should be defined by $ F_{\mu} = x^3 + y^3 + z^3 + 3 \mu xyz $. In that case, first observe that for $ \mu = 0 $ you get a nonsingular curve. So assume $ \mu \neq 0 $ and let $ [a:b:c] $ be a singular point with $ a \neq 0 $ WLOG. The equation $ 3a^2 = -3 \mu bc $ shows then that $ b,c $ are nonzero. We also have $ b^2 = - \mu ca $ and $ c^2 = - \mu ab $ and multiplying these three equalities and cancelling out $ a^2b^2c^2 $ gives $ 1 = - \mu^3 $.
Conversely, suppose $ \mu^3 = -1 $. Then we check straightaway that $ [-1: \mu : \mu ] $ is a singular point.