consider f(b) - f(a)

if f is a polynomial it's the sum of terms like n*x^p where n and p are integers and p >= 0. That difference, then, has terms that look like n*(b^p - a^p) where p > 0 and zero when p = 0.  Now, (b-a) is always a divisor of (b^p - a^p) regardless of p, so that difference must have (b-a) as a factor.

Using this with the first pair, f(b) - f(7) = 85 - 77 = 8 and so (b - 7) must divide 8. Also, b > 7 since we're given the order of the ages of the kids. (b-7), then, must be one of (1, 2, 4, 8).

Using this with the first and third gives f(c) - f(7) = 0 - 77 = -77, and so (c-7) must divide -77. (c-7) is positive, so it must be one of (1, 7, 11, 77).

Finally using this with the latter two ages gives f(c) - f(b) = 0 - 85 = -85 and so (c-b) must divide -85 and so must be one of (1, 5, 13, 85).

From the first pair, we have (b-7) in {1,2,4,8) and so b in {8, 9, 11, 15}

From the second pair we have (c-7) in {1,7,11,77} and so c in {8, 14, 18, 84}, but c cannot be 8 or else there'd be no possible age for b given that b >7 and b < c. So c is in {14, 18, 84}

Finally, the last pair demands (c-b) be one of {1, 5, 13, 85}. 85 isn't possible because c-b can't be > c. If the max difference is now 13, then c = 84 isn't possible because the min difference in that case is 69. The max difference remaining is 18 - 8 = 10, so (c-b) can't be 13. If c can now only be one of {14, 18} and (c-b) were 1 then b would have to be one of {13, 17} neither of which are possible. So (c-b) is 5, c is in {14, 18}, and b is one of {9,13}. Here at last there's a sole possibility, since b *can* be 9, and that's the solution.

a = 7, b = 9, and c = 14.

It's a pretty cool finding for a problem like this that we don't actually need to assume anything about the degree of the polynomial f. I'd not have expected that to be possible.