If `x+y+z=0` prove that `|a x b y c z c y a z b x b z c x a y|=x y z|a b cc a bb c a|`

Answer

Proved

Explanation

ATQ,
$$\left |\begin{array}{lll}x a & y b & z c \\ y c & z a & x b \\ z b & x c & y a\end{array}|=x y z|\begin{array}{ccc}a & b & c \\ c & a & b \\ b & c & a\end{array} \right |$$
$$\therefore \mathrm{LHS}=|\begin{array}{lll}\mathrm{xa} & \mathrm{yb} & \mathrm{zc} \\ \mathrm{yc} & \mathrm{za} & \mathrm{xb} \\ \mathrm{zb} & \mathrm{xc} & \mathrm{ya}\end{array}|$$
`=x a(z a \cdot y a-x b x c)-y b(y c \cdot y a-x b z b)+z c(y c \cdot x c-z a \cdot z b)`

`=x a(a^{2} y z-x^{2} b c)-y b(y^{2} a c-b^{2} x z)+z c(c^{2} x y-z^{2} a b)`

`=x y z a^{3}-x^{3} a b c-y 63 a b c+b^{3} x y z+c^{3} x y z-z^{3} a b c`

`=x y z(a^{3}+b^{3}+c^{3})-a b c(x^{3}+y^{3}+z^{3})`

`=x y z(a^{3}+b^{3}+c^{3})-a b c(3 x y z)`

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