Are there simple methods for calculating the determinant of symmetric matrices? I've seen that there are lots of exercises about determinants of symmetric matrices in my algebra books. Some are easy and others are a bit more twisted, but the basic problem is almost always the same. I have been trying to come up with a method to calculate these a bit more quickly, since—at least for me—they invariably end with a very ugly stream of numbers and letters.
For example I started with a $3\times 3$ matrix like this:
$$A= \begin{pmatrix}
a & b & c \\
b & a & b \\
c & b & a \end{pmatrix}$$
which looks fairly simple, but the best I could come up with for the determinant was:
$$2b^2(c-a)+a(a^2-c^2)
\quad
\text{ or }
\quad
a(a^2-2b^2-c^2)+2b^2c$$
These look horrific and absolutely not what anyone in his right mind would use. It goes without saying that I haven't even tried this with matrices bigger than $3\times 3$. Is there something I have been missing, or is there nothing to do about it?

In the more general case of a symmetric matrix
$$
A= \begin{pmatrix}
\color{brown}a & b & c \\
b & \color{brown}d & e \\
c & e & \color{brown}f \end{pmatrix}
$$
I quite like the expression for the determinant
$$
\det(A) = \color{brown}{adf} - (\color{brown}ae^2 + \color{brown}dc^2 + \color{brown}fb^2)+ 2bce \,.
$$
It's got a nice symmetry to it in terms of the diagonal entries. In your case where $a = d = f$ and $b = e$ it'll reduce to
$$
\color{brown}a^3 - \color{brown}a(b^2 + c^2 + b^2)  + 2bcb\,.
$$
The analogous presentation for the determinant of the $4\times 4$ matrix looks like:
$$\begin{align}
A = &\begin{pmatrix}
\color{brown}a & b & c & d\\
b & \color{brown}e & f &g \\
c & f & \color{brown}h &i \\
d&g&i&\color{brown}j
 \end{pmatrix}
\\[1em]
\implies\quad
\det(A) = \;\;
&\color{brown}{aehj}\\
-&(\color{brown}{ae}i^2
+\color{brown}{ah}g^2
+\color{brown}{aj}f^2
+\color{brown}{eh}d^2
+\color{brown}{ej}c^2
+\color{brown}{hj}b^2)
\\+&2(\color{brown}{a}fgi
+\color{brown}{e}cdi
+\color{brown}{h}bdg
+\color{brown}{j}bcf)
\\-&2(bfid + bcgi + cgfd)
+(bi)^2 + (cg)^2 + (df)^2
\end{align}$$
It's not beautiful, but again you can see some symmetry when the sum is sorted by how my factors of each summand are members of the diagonal. And again in your specific case where $a=e=h=j$ and $b=f=i$ and $c=g$ you get a slightly more memorable looking determinant:
$$
\color{brown}{a}^4
-\color{brown}a^2(3b^2+2c^2+d^2)
+4\color{brown}{a}(bdc + bcb)
-2(b^3d +b^2c^2+bc^2d) 
+b^4+c^4+b^2d^2
$$
In terms of a trick calculating these determinants, I was thinking of Leibniz's idea of a determinant as the sign-weighted sum over all products of an element from each row (or column) corresponding to an permutation of the symmetric group.
