, 2009

Solution 1. Rewrite the system as
$$\begin{cases} x^3 - 3x - 2 = -12(y - 4), \\ y^3 - 12y - 16 = 3(z - 6), \\ z^3 - 27z - 54 = 27(x - 2). \end{cases}$$
Then factor the left sides to obtain
$$\begin{cases} (x+1)^2(x-2) = -12(y-4), \\ (y+2)^2(y-4) = 3(z-6), \\ (z+3)^2(z-6) = 27(x-2). \end{cases}$$
Multiplying the three equations and moving everything to one side we obtain
$$(x-2)(y-4)(z-6)[((x+1)(y+2)(z+3))^2 + 972] = 0.$$
Since squares are nonnegative, either $x = 2$ or $y = 4$ or $z = 6$. But by examining the above system we see that any of these equalities implies the other two. This gives the unique solution $(x, y, z) = (2, 4, 6)$.

Solution 2 (By Gabriel Carroll). Write
$$\begin{align*} F(x) &= \frac{x^3 - 3x - 50}{-12}, \\ G(y) &= \frac{y^3 - 12y + 2}{3}, \\ H(z) &= \frac{z^3 - 27z}{27}. \end{align*}$$
Then $y = F(x)$, $z = G(y)$, $x = H(z)$. Because $-4F(x)' = x^2 - 1$, $F(x)$ is increasing on the interval $[-1, 1]$ (and decreasing otherwise), with $F(-1) = F(2) = 4$. Likewise, we can show that $G(x)$ is decreasing on $[-2, 2]$ (and increasing otherwise), with $G(-2) = G(4) = 6$; and $H(x)$ is decreasing on $[-3, 3]$ (and increasing otherwise), with $H(-3) = H(6) = 2$.
If $x > 2$ then we get $y = F(x) < F(2) = 4$. Consequently, $z = G(y) \le 6$ and so $x = H(z) \le 2$, which is a contradiction. Likewise, if $x < 2$ then $y \ge 4$, $z \ge 6$, and $x \ge 2$, which is a contradiction again. Hence $x = 2$, leading to the unique solution $(x, y, z) = (2, 4, 6)$.