value of abc in cubic equation

Without loss of generality we may assume c≤a≤b.
Let c = a-n & b = a+m where m,n≥0.

Using elementary algebra we can find that,

3b²+3c²-a³ ≥ 3a²+3c²-b³

This is because-

3(a+m)² + 3(a-n)² - a³ ≥ 3a²+3(a-n)²-(a+m)³
where equality holds only when m=n=0.

Since in the above equations the RHS and the LHS of the inequality are both equal to 25.

m=n=0 or a=b=c

Therefore,
3b²+3c²-a³
=3a²+3a²-a³ =25
or, a=5

Therfore a=b=c=5