, 2003

Solution:

The answer is (C). Given a prime $p$ we denote by $x_{p}, y_{p}$ and $z_{p}$ the exponent (possibly zero) with which $p$ appears in the factorization of $x$, of $y$ and of $z$ respectively; in the same way we fix $a_{p}, b_{p}$ and $c_{p}$. The hypothesis of (C) is equivalent to saying that $x_{p}, y_{p}, z_{p}$ are multiples of 3 for every prime $p$. We thus know that 3 divides the numbers $(a_{p} + b_{p}), (a_{p} + c_{p})$ and $(b_{p} + c_{p})$ and therefore 3 also divides:
$$(a_{p} + b_{p}) + (a_{p} + c_{p}) - (b_{p} + c_{p}) = 2 a_{p}$$
so $a_{p}$ is divisible by 3. In exactly the same way one verifies that $b_{p}$ and $c_{p}$ are also divisible by 3, this holds for every $p$ and therefore $a, b, c$ are cubes.

We give below counterexamples for the other answers:
(A) $a = b = c = 2$;
(B) $a = b = 2, c = 1$;
(D) $a = b = 10, c = 1$.