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 divided by $(x+y+z)$ and hence find the quotient.

Answer 1

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

Applying $C_1\to C_1+C_2+C_3$

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

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

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

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

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

$\Rightarrow \Delta ={(x+y+z)}^2(xy+yz+zx-x^2-y^2-z^2)[(y-z)(z-y)-(z-x)(y-x)-0+0]$

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

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

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