Let a, b, c, d be complex numbers satisfying $a+b+c+d=a^3+b^3+c^3+d^3=0$. Prove that a pair of the a, b, c, d must add up to 0

Question

When doing this I tried using the identity $x^3+y^3+z^3=3xyz$ if $x+y+z=0$

I take $x=a$, $y=b$, and $z=c+d$

So $a+b+(c+d)=0$

$a^3+b^3+(c+d)^3=3ab(c+d)$

$a^3+b^3+c^3+d^3+3cd(c+d)=3ab(c+d)$

$(a^3+b^3+c^3+d^3)=3ab(c+d)-3cd(c+d)$

$0=3(c+d)(ab-cd)$

So I got $c+d=0$ or $ab=cd$

If $c+d=0$, the statement is true, but how about if it's $ab=cd$?