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?


Edit (July 2021): As suggested in the comment, the answer here calculated the determinant of
$$\begin{pmatrix}
\ a & b & c \\
b & c & a \\
c & a & b \end{pmatrix},$$
instead of the one in the post.

Original answer:
do R1 --> R1+R2+R3
take out $(a+b+c)$
you will end up with
$$=(a+b+c)\begin{pmatrix}
\ 1 & 1 & 1 \\
b & c & a \\
c & a & b \end{pmatrix}$$
c1 --> c1-c3
c2 --> c2-c3
$$=(a+b+c)\begin{pmatrix}
\ 0 & 0 & 1 \\
b-a & c-a & a \\
c-b & a-b & b \end{pmatrix}$$
expanding along R1:
$$=(a+b+c)(a^2 +b^2 + c^2 -ab-bc-ca)$$
