12.15. Solve the equation in integers

$$
x^{3}-2 y^{3}-4 z^{3}=0
$$
12.15. Answer: $x=y=z=0$.

Let $x^{3}-2 y^{3}-4 z^{3}=0$, where $x, y, z$ are integers. Then the number $x$ is even. After substituting $x=2 x_{1}$, we get the equation $8 x_{1}^{3}-2 y^{3}-4 z^{3}=0$. Dividing by 2: $4 x_{1}^{3}-y^{3}-2 z^{3}=0$. Therefore, the number $y$ is even. After substituting $y=2 y_{1}$, we get the equation $4 x_{1}^{3}-8 y_{1}^{3}-2 z^{3}=0$. Dividing by 2 again: $2 x_{1}^{3}-4 y_{1}^{3}-z^{3}=0$. Therefore, the number $z$ is even. After substituting $z=2 z_{1}$, we get the equation $x_{1}^{3}-2 y_{1}^{3}-4 z_{1}^{3}=0$, which has the same form as the original equation. Therefore, we can again prove that the numbers $x_{1}, y_{1}, z_{1}$ are even, and so on. But this is only possible if $x=y=z=0$.