Question

Prove that $\left|\begin{array}{ccc}{yz-x}^2& {zx-y}^2& xy-z^2\\ {zx-y}^2& xy-z^2& {yz-x}^2\\ {xy-z}^2& {yz-x}^2& {zx-y}^2\end{array}\right|$ is divisible by $(x+y+z)$ and hence find the quotient

Answer 1

Let $\Delta =\left|\begin{array}{ccc}yz-x^2& zx-y^2& xy-2^2\\ zx-y^2& xy-2^2& yz-x^2\\ xy-2^2& yz-x^2& zx-y^2\end{array}\right|$

Adding $C_1\to C_1+C_2+C_3$, we get

$\Delta =\left|\begin{array}{ccc}xy+yz+zx-x^2-y^2-z^2& zx-y^2& xy-2^2\\ xy+yz+zx-x^2-y^2-z^2& xy-2^2& yz-x^2\\ xy+yz+zx-x^2-y^2-z^2& yz-x^2& zx-y^2\end{array}\right|$

$\Delta =(xy+yz+zx-x^2-y^2-z^2)\left|\begin{array}{ccc}1& zx-y^2& xy-2^2\\ 1& xy-2^2& yz-x^2\\ 1& yz-x^2& zx-y^2\end{array}\right|$

Applying $R_2\to R_2-R_1$ and $R_3\to R_3-R_1$ we get

$\Delta =(xy+yz+zx-x^2-y^2-z^2)\left|\begin{array}{ccc}1& zx-y^2& xy-2^2\\ 0& (x+y+z)(y-z)& (x+y+z)(z-x)\\ 0& (x+y+z)(y-x)& (x+y+z)(z-y)\end{array}\right|$

$\Delta =(xy+yz+zx-x^2-y^2-z^2){(x+y+z)}^2\left|\begin{array}{ccc}1& zx-y^2& xy-2^2\\ 0& (y-z)& (z-x)\\ 0& (y-x)& (z-y)\end{array}\right|$

$\Delta =(xy+yz+zx-x^2-y^2-z^2){(x+y+z)}^2[(y-z)(z-y)-(z-x)(y-x)]$

$\Delta =(x^2+y^2+z^2-xy-yz-zx){(x+y+z)}^2$

so $\Delta$ is divisible by $(x+y+z)$

The quotient when $\Delta$ is divisible by $(x+y+z)$ is $(x^2+y^2+z^2-xy-yz-zx)(x+y+z)$