Infinite descent

Note that if \(x, y, z\) are rational numbers satisfying the given equation, then \(dx, dy, dz\) also satisfy the equation for any positive integer \(d\). Hence, it suffices to prove that there are no integer solutions to the given equation other than the trivial solution \(x = y = z = 0\). We claim that if \((a, b, c)\) are integers satisfying \[ x^3 + 3y^3 + 9z^3 = 9xyz, \] then \(3\) divides \(a\), and \((b, c, a/3)\) also satisfies the above equation. Indeed, if we assume that if \((a, b, c)\) are integers satisfying \[ x^3 + 3y^3 + 9z^3 = 9xyz, \] then note that \(3\) divides \(a^3\). Since \(3\) is a prime, it follows that \(3\) divides \(a\). Using \[ a^3 + 3b^3 + 9c^3 = 9abc, \] we obtain \[ b^3 + 3c^3 + 9 \left( \frac{a}{3} \right)^3 = 9 bc \left( \frac{a}{3} \right) . \] This completes the proof of the claim.

Let \((x, y, z)\) be a non-trivial integer solution to the given equation with \(\vert x \vert + \vert y \vert + \vert z \vert\) minimum. Note that \(x\) is nonzero, otherwise, \(y, z\) satisfy \(y^3 + 3 z^3 = 0\), which is impossible since \(y, z\) are integers. By the above claim, it follows that \(y, z, x/3\) is also a solution to the given equation. Using that \(x\) is nonzero, we obtain \[ \vert y \vert + \vert z \vert + \left \vert \frac{x}{3} \right \vert < \vert x \vert + \vert y \vert + \vert z \vert , \] which contradicts the minimality of \(\vert x \vert + \vert y \vert + \vert z \vert\). This shows that there are no non-trivial integer solutions to the given equation.