We will show that there is no solution other than the trivial $a = b = c = 0$. For any other triple, we can write $a = dx$, $b = dy$, $c = dz$, where $\gcd(x, y, z) = 1$ and $d$ is a positive integer. After dividing out $d$, we obtain the same equation but for $x$, $y$, and $z$. We will demonstrate that each of $x$, $y$, and $z$ must be divisible by $3$, which leads to a contradiction.

Assume first that none of $x$, $y$, or $z$ is divisible by $3$. Then $x^2 \equiv y^2 \equiv z^2 \equiv 1 \pmod{3}$, which implies that
$$x + y + z \equiv xyz + y + z + x \pmod{3},$$
meaning $xyz \equiv 3 \pmod{3}$, a contradiction. Thus, we may assume that, for example, $z$ is divisible by $3$. Then $3$ also divides $x^3 + y^3 - 25x^2y$. Note that we always have $3 \mid x^3 - x$, $3 \mid y^3 - y$ thus we get $3 \mid x + y(1 - x^2)$. Now, if $3$ does not divide $x$, then it would not divide $y(1 - x^2)$, which is a contradiction since $3 \mid 1 - x^2$. Therefore, $3 \mid x$, and consequently also $y$, as required. $\square$