905. Solve the equation

$$
x^{3}=3 y^{3}+9 z^{3}
$$

in non-negative integers $x, y$ and $z$.
$\triangle$ From the equation, it is clear that the number $x$ is divisible by 3: $x=3 x_{1}$, where $x_{1}$ is a non-negative integer. We obtain:

$$
27 x_{1}^{3}=3 y^{3}+9 z^{3}, \quad 9 x_{1}^{3}=y^{3}+3 z^{3}
$$

From the last equation, it follows that $y$ is divisible by 3: $y=3 y_{1}$, where $y_{1}$ is a non-negative integer. Then

$$
9 x_{1}^{3}=27 y_{1}^{3}+3 z^{3}, \quad 3 x_{1}^{3}=9 y_{1}^{3}+z^{3}
$$

Now $z$ is also divisible by 3: $z=3 z_{1}$ ($z_{1}$ - a non-negative integer). We obtain:

$$
x_{1}^{3}=3 y_{1}^{3}+9 z_{1}^{3}
$$

Notice that we have obtained an equation of the same form as the original one. Therefore, from this equation, we can similarly find that the numbers $x_{1}, y_{1}$, and $z_{1}$ are divisible by 3. Consequently, the numbers $x, y$, and $z$ are divisible by $3^{2}$.

Similarly, we find that the numbers $x, y$, and $z$ are divisible by $3^{3}$, then by $3^{4}$, and generally by $3^{n}$, where $n$ is any natural number. However, the only non-negative integer (and generally any integer) that has this property is zero. Moreover, the values

$$
x=0, \quad y=0, \quad z=0
$$

satisfy the equation.

Answer: $(0 ; 0 ; 0)$.

The method of solution we used here is called the method of infinite descent. It was first introduced into number theory by P. Fermat in the 17th century.