Determine all integers $a, b, c$ satisfying the identities:

$$
\begin{gathered}
a+b+c=15 \\
(a-3)^{3}+(b-5)^{3}+(c-7)^{3}=540
\end{gathered}
$$
Solution II: We use the substitution $a-3=x, b-5=y, c-7=z$.

Now, equations are transformed to:

$$
\begin{aligned}
x+y+z & =0 \\
x^{3}+y^{3}+z^{3} & =540 .
\end{aligned}
$$

Substituting $z=-x-y$ in the second equation, we get:

$$
-3 x y^{2}-3 x^{2} y=540
$$

or

$$
x y(x+y)=-180
$$

or

$$
x y z=180 \text {. }
$$

Returning to the starting problem we have:

$$
(a-3)(b-5)(c-7)=180
$$