If $A+B+C=0$, then prove that the value of the determinant is $0$. I'll state the question from my textbook below:

If $A+B+C=0$, the prove that $\begin{vmatrix}1 & \cos C & \cos B \\ \cos C & 1 & \cos A \\ \cos B & \cos A & 1 \end{vmatrix} = 0$.

This is how I tried solving the problem:
$LHS = \begin{vmatrix}1 & \cos C & \cos B \\ \cos C & 1 & \cos A \\ \cos B & \cos A & 1 \end{vmatrix}$
$= 1(1- \cos^2 A) - \cos C (\cos C - \cos A \cos B) + \cos B (\cos A \cos C - \cos B)$
$= 1 + 2 \cos A \cos B \cos C - (\cos^2 A +\cos^2 B + \cos^2 C)$
I don't know how to proceed further. I tried using the fact that $A+B+C=0$ but it didn't lead to anything I could solve. I don't know where is it supposed to be used. 
Also, I read a solution to this problem somewhere in which the term $(\cos^2 A +\cos^2 B + \cos^2 C)$ was replaced by $1 + 2 \cos A \cos B \cos C$ as $A+B+C=0$. Are these two terms equal for the given condition? Also, is there a way to prove the statement without using this fact?
Any help would be appreciated.

Using Prove that $\cos (A + B)\cos (A - B) = {\cos ^2}A - {\sin ^2}B$,
$$\cos^2A+\cos^2B+\cos^2C =1+\cos^2A-\sin^2B+\cos^2C$$
$$=1+\cos(A+B)\cos(A-B)+\cos^2C$$
$$=1-\cos C\cos(A-B)+\cos^2C\text{ as }\cos(A+B)=\cos(\pi-C)=?$$
$$=1-\cos C[\cos(A-B)-\cos C]$$
$\cos(A-B)-\cos C=\cos(A-B)+\cos(A+B)=?$
