Simplified method for symmetric matrix determinants

There's not a whole lot you can do to simplify that anymore.

In any case, what you've written down is a special case of a symmetric matrix. In general a symmetric $3 \times 3$ matrix will have the form:

$$A= \begin{pmatrix} \ a & b & c \\ b & d & e \\ c & e & f \end{pmatrix}$$

which has a determinant of $a(df-e^2) + b(ce-bf) + c(be-dc)$. Even worse-looking.

The only time it really gets a lot simpler is if you have zeroes in there. The simplest way to calculate is not to calculate.

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:
$$=(a+b+c)(a^2 +b^2 + c^2 -ab-bc-ca)$$

What is “elegant”? If $a \ne 0$ and $M_{i,j}$ is the $(i, j)$ minor of a matrix $M$ (see Wikipedia), then the determinant of the matrix $M$ you present is $(M_{3,3}^2 - M_{1,3}^2)/a$. This is arguably more “elegant”.