You had a nice symmetric matrix with all diagonal elements equal. That's an especially simple form as others already noted of course. It's maybe worth seeing that your equation is analytically not so crazy as all that and it can be solved to some extent to cough up some useful stuff.
Testing for a zero determinant
Look at what always happens when c=a. Disaster for invertibility. The determinant for that kind of a matrix must always be zero. When you get an equation like this for a determinant, set it equal to zero and see what happens! Those are by definition a description of all your singular matrices.
That's arguably the most useful test to think about straight up because you only have to worry about calculating the value of the determinant if it isn't zero. Another way of saying if your symmetric matrix has all diagonal entries the same as in your original post, you can use your equation to test if that determinant is zero without actually calculating the determinant.
This function might still look awful but it has zeros in other places than the trivial case where a=c. This is also zero where,
Note that: (a2−c2)=(a+c)(a-c) meaning(a-c) can be factored from LHS and RHS,
Which solves for c,
c = (2b2-a2)/a
Or if you prefer b,
b = sqrt(a2+ac)
If either c=a or c = (2b2-a2)/a hold then you're absolutely guaranteed a zero determinant and you don't have to actually work out the determinant from computing anything but
(2b2-a2)/a and comparing that to the value of c. Of course this only holds for matrices of the form you posted with all main diagonal elements the same.
Determinants by the extended matrix/diagonals method
If you do want a neat brute force method for working out determinants and in a way that makes it almost impossible to go wrong just because it is so organised, there's the so-called American method. This is hard to beat for simplicty but it does involve some redundancy.
Suppose you have,
Now switch to augmented matrix notation... take the first two columns and do something that looks utterly redundant but which will end up being really simplifying. Re-write the first two columns all over again to the right of the original matrix, to get something like,
If you look at the diagonals which are to the right of the main diagonal and parallel to it, if you were to take the product of any three terms in those three neighbouring diagonals well, just look at them, they are all the positive terms appearing in the determinant, now regrouped into diagonals.
So that's (1 x 5 x 9) + (2 x 6 x 7) + (3 x 4 x 8)=(45)+(84)+(96)=225 and those are the positive terms.
From that you take off all the terms which would be subtracted and lo, you can locate on the three diagonals which are running the other way.
So we'll have to subtract (3 x 5 x 7) + (1 x 6 x 8) + (2 x 4 x 9) = 105 + 48 + 72=225
So the determinant in this case is 225-225 = 0.
It never fails.
This method just regroups the positive and negative terms in the determinant into diagonals. So my own preference for matrices where there are no easy rows or columns to work with, where a matrix isn't very sparse and you can't use other methods to save a lot of time, this more long-winded looking method works pretty rapidly for determinants of any dimension and it is almost impossible to mess this up.