Prove that $\begin{vmatrix} xa&yb&zc\\ yc&za&xb\\ zb&xc&ya\\ \end{vmatrix}=xyz\begin{vmatrix} a&b&c\\ c&a&b\\ b&c&a\\ \end{vmatrix}$ if $x+y+z=0$ 
If $x+y+z=0$, then prove that
  $$
\begin{vmatrix}
xa&yb&zc\\
yc&za&xb\\
zb&xc&ya\\
\end{vmatrix}=xyz\begin{vmatrix}
a&b&c\\
c&a&b\\
b&c&a\\
\end{vmatrix}
$$

I can do it by Sarrus' law but how can I prove it by matrix operations without actually expanding the determinant ?
My Attempt
$$
\begin{vmatrix}
xa&yb&zc\\
yc&za&xb\\
zb&xc&ya\\
\end{vmatrix}=xyz\begin{vmatrix}
a&\frac{yb}{x}&\frac{zc}{x}\\
c&\frac{za}{y}&\frac{xb}{y}\\
b&\frac{xc}{z}&\frac{ya}{z}\\
\end{vmatrix}=xyz\begin{vmatrix}
a&-b-\frac{zb}{x}&-c-\frac{yc}{x}\\
c&-a-\frac{xa}{y}&-b-\frac{zb}{y}\\
b&-c-\frac{yc}{z}&-a-\frac{xa}{z}\\
\end{vmatrix}=xyz\begin{vmatrix}
a&b+\frac{zb}{x}&c+\frac{yc}{x}\\
c&a+\frac{xa}{y}&b+\frac{zb}{y}\\
b&c+\frac{yc}{z}&a+\frac{xa}{z}\\
\end{vmatrix}=xyz\begin{vmatrix}
a&b&c+\frac{yc}{x}\\
c&a&b+\frac{zb}{y}\\
b&c&a+\frac{xa}{z}\\
\end{vmatrix}+xyz\begin{vmatrix}
a&\frac{zb}{x}&c+\frac{yc}{x}\\
c&\frac{xa}{y}&b+\frac{zb}{y}\\
b&\frac{yc}{z}&a+\frac{xa}{z}\\
\end{vmatrix}\\
=xyz\bigg(\begin{vmatrix}
a&b&c\\
c&a&b\\
b&c&a\\
\end{vmatrix}+\begin{vmatrix}
a&b&\frac{yc}{x}\\
c&a&\frac{zb}{y}\\
b&c&\frac{xa}{z}\\
\end{vmatrix}+\begin{vmatrix}
a&\frac{zb}{x}&c\\
c&\frac{xa}{y}&b\\
b&\frac{yc}{z}&a\\
\end{vmatrix}+\begin{vmatrix}
a&\frac{zb}{x}&\frac{yc}{x}\\
c&\frac{xa}{y}&\frac{zb}{y}\\
b&\frac{yc}{z}&\frac{xa}{z}\\
\end{vmatrix}\bigg)\\
$$
I need to prove that the sum of last three terms is zero.
$$
\begin{vmatrix}
a&b&\frac{yc}{x}\\
c&a&\frac{zb}{y}\\
b&c&\frac{xa}{z}\\
\end{vmatrix}+\begin{vmatrix}
a&\frac{zb}{x}&c\\
c&\frac{xa}{y}&b\\
b&\frac{yc}{z}&a\\
\end{vmatrix}+\begin{vmatrix}
a&\frac{zb}{x}&\frac{yc}{x}\\
c&\frac{xa}{y}&\frac{zb}{y}\\
b&\frac{yc}{z}&\frac{xa}{z}\\
\end{vmatrix}=\begin{vmatrix}
a&b&\frac{yc}{x}\\
c&a&\frac{zb}{y}\\
b&c&\frac{xa}{z}\\
\end{vmatrix}+\begin{vmatrix}
a&\frac{zb}{x}&\frac{-zc}{x}\\
c&\frac{xa}{y}&\frac{-xb}{y}\\
b&\frac{yc}{z}&\frac{-ya}{z}\\
\end{vmatrix}\\
$$
Solution by expansion
$$
\Delta=\begin{matrix}
xa&yb&zc&xa&yb\\
yc&za&xb&yc&za\\
zb&xc&ya&zb&xc\\
\end{matrix}=xyz(a^3+b^3+c^3)-abc(x^3+y^3+z^3)
$$
We have $x^3+y^3+z^3-3xyz=(x+y+z)(x^2+y^2+z^2-xy-yz-zx)=0$ as $x+y+z=0$. Thus, $x^3+y^3+z^3=3xyz$
$$
\Delta=xyz(a^3+b^3+c^3)-abc(3xyz)=xyz(a^3+b^3+c^3-3abc)\\
=xyz\bigg[ a\big(a^2-bc\big)-b\big(ac-b^2\big)+c\big(c^2-ab\big) \bigg]\\
=xyz.\bigg[a\begin{vmatrix}
a&b\\
c&a\\
\end{vmatrix}-b\begin{vmatrix}
c&b\\
b&a\\
\end{vmatrix}+c\begin{vmatrix}
c&a\\
b&c\\
\end{vmatrix}\bigg]
=xyz\begin{vmatrix}
a&b&c\\
c&a&b\\
b&c&a\\
\end{vmatrix}
$$
 A: Possible direction, though not a proof. 
Do note that curiously, the identity is only true for $n=3$, in higher dimensions it doesn't seem to work. Note also that the matrices do not actually have to be positive semi-definite matrices.
This result may be related to Oppenheim's inequality for Hadamard products, namely if $A,X$:
$$A=\begin{pmatrix} a&b&c\\ c&a&b\\ b&c&a\\ \end{pmatrix},\quad X=\begin{pmatrix} x&y&z\\ y&z&x\\ z&x&y\\ \end{pmatrix}$$
Are two positive semi definite matrices, then:
$$\det(A\circ X)\ge (\det A)\prod_i x_{ii}$$
With equality if $A\circ X$ is singular. In your case, you have a circulant and anti-circulant matrices $A,X$ and the above theorem states that:
$$\det(A\circ X)\ge xyz\det(A)$$
$$\begin{vmatrix} xa&yb&zc\\ yc&za&xb\\ zb&xc&ya\\ \end{vmatrix} \ge xyz\begin{vmatrix} a&b&c\\ c&a&b\\ b&c&a\\ \end{vmatrix}$$
You are looking for equality and we of course don't know that either $A$ or $X$ is actually PSD. Moreover we haven't used the condition $x+y+z=0$. Nevertheless, the overall form direction may be one worth pursuing.
Further Reading: The equality cases for the inequalities of Oppenheim and Schur for positive semi-definite matrices, Czechoslovak Mathematical Journal, Vol. 59 (2009), No. 1, 197–206
A: I don't think it is possible without expanding the determinants. The determinant on the right is $a^3+b^3+c^3-3abc = (a+b+c)(a^2+b^2+c^2-ab-ac-bc)$. The determinant on the left is $(a^3+b^3+c^3)xyz-abc(x^3+y^3+z^3)$. Subtracting the left side from the right side is equal to $abc(x^3+y^3+z^3-3xyz)=abc(x+y+z)(x^2+y^2+z^2-xy-xz-yz)$. The result follows.
