Let $$\Delta=\left|\begin{array}{lll} b c-a^2 & c a-b^2 & a b-c^2 \\ c a-b^2 & a b-c^2 & b c-a^2 \\ a b-c^2 & b c-a^2 & c a-b^2 \end{array}\right|$$

$\begin{array}{r}=\left|\begin{array}{lll}b c-a^2-c a+b^2 & c a-b^2-a b+c^2 & a b-c^2 \\ c a-b^2-a b+c^2 & a b-c^2-b c+a^2 & b c-a^2 \\ a b-c^2-b c+a^2 & b c-a^2-c a+b^2 & c a-b^2\end{array}\right| \\ \quad\left[\because C_1 \rightarrow C_1-C_2 \text { and } C_2 \rightarrow C_2-C_3\right]\end{array}$

$$\begin{aligned} & =\left|\begin{array}{lll} (b-a)(a+b+c) & (c-b)(a+b+c) & a b-c^2 \\ (c-b)(a+b+c) & (a-c)(a+b+c) & b c-a^2 \\ (a-c)(a+b+c) & (b-a)(a+b+c) & c a-b^2 \end{array}\right| \\ & =(a+b+c)^2\left|\begin{array}{lll} b-a & c-b & a b-c^2 \\ c-b & a-c & b c-a^2 \\ a-c & b-a & c a-b^2 \end{array}\right|\quad \text { [taking }(a+b+c) \text { common from } C_1 \text { and } C_2 \text { each] } \end{aligned}$$

$\begin{array}{r}=(a+b+c)^2\left|\begin{array}{ccc}0 & 0 & a b+b c+c a-\left(a^2+b^2+c^2\right) \\ c-b & a-c & b c-a^2 \\ a-c & b-a & c a-b^2\end{array}\right| \\ \quad\left[\because R_1 \rightarrow R_1+R_2+R_3\right]\end{array}$

$$\begin{aligned} &\text { Now, expanding along } R_1 \text {, }\\ &\begin{aligned} & \left.=(a+b+c)^2\left[a b+b c+c a-\left(a^2+b^2+c^2\right)\right](c-b)(b-a)-(a-c)^2\right] \\ & =(a+b+c)^2\left(a b+b c+c a-a^2-b^2-c^2\right) \\ & \quad \quad\left(c b-a c-b^2+a b-a^2-c^2+2 a c\right) \\ & =(a+b+c)^2\left(a^2+b^2+c^2-a b-b c-c a\right) \\ & \quad\left(a^2+b^2+c^2-a c-a b-b c\right) \\ & =\frac{1}{2}(a+b+c)\left[(a+b+c)\left(a^2+b^2+c^2-a b-b c-c a\right)\right] \\ & \quad\left[(a-b)^2+(b-c)^2+(c-a)^2\right] \\ & =\frac{1}{2}(a+b+c)\left(a^3+b^3+c^3-3 a b c\right)\left[(a-b)^2+(b-c)^2+(c-a)^2\right] \end{aligned} \end{aligned}$$

$$\begin{aligned} &\text { Hence, given determinant is divisible by }(a+b+c) \text { and quotient is }\\ &\left(a^3+b^3+c^3-3 a b c\right)\left[(a-b)^2+(b-c)^2+(c-a)^2\right] \end{aligned}$$

$$\begin{aligned} &\text { Since, } x+y+z=0 \text {, also we have to prove }\\ &\begin{aligned} \left|\begin{array}{ccc} x a & y b & z c \\ y c & z a & x b \\ z b & x c & y a \end{array}\right| & =x y z\left|\begin{array}{ccc} a & b & c \\ c & a & b \\ b & c & a \end{array}\right| \\ \therefore\quad \text { LHS } & =\left|\begin{array}{ccc} x a & y b & z c \\ y c & z a & x b \\ z b & x c & y a \end{array}\right| \end{aligned} \end{aligned}$$

$$\begin{aligned} & =x a(z a \cdot y a-x b \cdot x c)-y b(y c \cdot y a-x b \cdot z b)+z c(y c \cdot x c-z a \cdot z b) \\ & =x a\left(a^2 y z-x^2 b c\right)-y b\left(y^2 a c-b^2 x z\right)+z c\left(c^2 x y-z^2 a b\right) \\ & =x y z a^3-x^3 a b c-y^3 a b c+b^3 x y z+c^3 x y z-z^3 a b c \\ & =x y z\left(a^3+b^3+c^3\right)-a b c\left(x^3+y^3+z^3\right) \\ & =x y z\left(a^3+b^3+c^3\right)-a b c(3 x y z) \\ & \quad \quad\left[\because x+y+z=0 \Rightarrow x^3+y^3+z^3-3 x y z\right] \\ & =x y z\left(a^3+b^3+c^3-3 a b c\right)\quad\text{.... (i)} \end{aligned}$$

Now, $$ \mathrm{RHS}=x y z\left|\begin{array}{lll} a & b & c \\ c & a & b \\ b & c & a \end{array}\right|=x y z\left|\begin{array}{lll} a+b+c & b & c \\ a+b+c & a & b \\ a+b+c & c & a \end{array}\right| \quad\left[\because C_1 \rightarrow C_1+C_2+C_3\right]$$

$=x y z(a+b+c)\left|\begin{array}{lll}1 & b & c \\ 1 & a & b \\ 1 & c & a\end{array}\right| \quad\left[\operatorname{taking}(a+b+c)\right.$ common from $\left.C_1\right]$

$\begin{aligned} &=x y z(a+b+c)\left|\begin{array}{ccc}0 & b-c & c-a \\ 0 & a-c & b-a \\ 1 & c & a\end{array}\right| \\ & {\left[\because R_1 \rightarrow R_1-R_3 \text { and } R_2 \rightarrow R_2-R_3\right] }\end{aligned}$

$$\begin{aligned} &\text { Expanding along } \mathrm{C}_1 \text {, }\\ &\begin{aligned} & =x y z(a+b+c)[1(b-c)(b-a)-(a-c)(c-a)] \\ & =x y z(a+b+c)\left(b^2-a b-b c+a c+a^2+c^2-2 a c\right) \\ & =x y z(a+b+c)\left(a^2+b^2+c^2-a b-b c-c a\right) \\ & =x y z\left(a^3+b^3+c^3-3 a b c\right)\quad\text{.... (ii)} \end{aligned} \end{aligned}$$

$$\begin{aligned} &\text { From Eqs. (i) and (ii), }\\ &\mathrm{LHS}=\mathrm{RHS}\\ &\Rightarrow \quad\left|\begin{array}{lll} x a & y b & z c \\ y c & z a & x b \\ z b & x c & y a \end{array}\right|=x y z\left|\begin{array}{ccc} a & b & c \\ c & a & b \\ b & c & a \end{array}\right|\quad\text{Hence proved.} \end{aligned}$$