Solution:

First suppose none of them are $0$. Let the common value of the four expressions be $k$, and let $a b c d = P$. Then for $x \in \{a, b, c, d\}$,
$$x^{3} - \frac{P}{x} = k \Longrightarrow x^{4} - k x - P = 0$$
However, Vieta's tells us $a b c d = -P$, meaning $P = -P$, so $P = 0$, a contradiction.

Now if $a = 0$, then $-b c d = b^{3} = c^{3} = d^{3}$. Then without loss of generality $b = x$, $c = x \omega$, and $d = x \omega^{2}$. But then $-b c d = -x^{3} \neq x^{3}$, a contradiction.

Thus, there do not exist distinct complex numbers satisfying the equation.

Solution:

Subtracting the first two equations and dividing by $a-b$ gives $a^{2} + b^{2} + a b + c d = 0$. Similarly, $c^{2} + d^{2} + a b + c d = 0$. So, $a^{2} + b^{2} = c^{2} + d^{2}$. Similarly, $a^{2} + c^{2} = b^{2} + d^{2}$. So, $b^{2} = c^{2}$. Similarly, $a^{2} = b^{2} = c^{2} = d^{2}$. Now by Pigeonhole, two of these 4 must be the same.