USA Team Selection Test

To find all triples $(x, y, z)$ of positive integers such that $x \leq y \leq z$ and
$$x^3(y^3 + z^3) = 2012(xyz + 2),$$
we proceed as follows:

First, note that $2012 \cdot 2 = 2^3 \cdot 503$. Taking the equation modulo $x$, we see that $x \mid 2012$. Therefore, $x$ can be $1, 2, 4, 503, 1006, 2012$. We will consider the feasible values of $x$ under the constraint $x \leq y \leq z$.

### Case 1: $x = 1$
The equation becomes:
$$y^3 + z^3 = 2012(yz + 2).$$
Factoring the left-hand side, we get:
$$(y + z)(y^2 - yz + z^2) = 2012(yz + 2).$$
For $503 \mid y + z$ or $503 \mid y^2 - yz + z^2$, we apply the lemma that $503 \nmid y^2 - yz + z^2$ for positive integers $y$ and $z$ not both divisible by 503. Therefore, $503 \mid y + z$. Let $y + z = 503n$.

Rewriting the equation:
$$n(y^2 - yz + z^2) = 4(yz + 2).$$
If $n \geq 4$, then $(z - y)^2 \leq 2$, implying $z = y$ or $z = y + 1$. Both cases lead to contradictions. Thus, $n < 4$.

Since $y$ and $z$ have the same parity, $n = 2$ is required, leading to $y + z = 1006$. This results in:
$$(y + z)^2 = 1006^2 = 5yz + 4,$$
which is a contradiction modulo 5. Hence, no solutions exist for $x = 1$.

### Case 2: $x = 2$
The equation becomes:
$$y^3 + z^3 = 503(yz + 1).$$
Letting $y + z = 503n$, we rewrite this as:
$$n(y^2 - yz + z^2) = yz + 1.$$
If $n \geq 3$, then $n(z - y)^2 = 1 + (1 - n)yz < 0$, a contradiction.

For $n = 1$, we have:
$$y^2 - yz + z^2 = yz + 1,$$
or $(z - y)^2 = 1$, leading to $z = y + 1$. Given $y + z = 503$, we find $y = 251$ and $z = 252$. Verifying, $(x, y, z) = (2, 251, 252)$ satisfies the original equation.

For $n = 2$, we get:
$$2(y^2 - yz + z^2) = yz + 1,$$
or $2(z - y)^2 = 1 - yz$, forcing $y = z = 1$, which contradicts $y + z = 1006$.

Thus, the only solution is $(x, y, z) = (2, 251, 252)$.

The answer is: $\boxed{(2, 251, 252)}$.