Prove that `|(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)|`is divisible by (*x* + *y* + *z*) and hence find the quotient.

#### Solution

Let ∆= `|(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)|`

Applying C_{1}→C_{1}+C_{2}+C_{3} , we get

∆= `|(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)|`

⇒∆=(xy+yz+zx−x^{2}−y^{2}−z^{2})`|(1,zx-y^2,xy-z^2),(1,xy-z^2,yz-x^2),(1,yz-x^2,zx-y^2)|`

Applying R_{2}→R_{2}−R_{1} and R_{3}→R_{3}−R_{1}, we get

∆=(xy+yz+zx−x^{2}−y^{2}−z^{2})`|(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))|`

⇒∆=(x+y+z)^{2}(xy+yz+zx−x^{2}−y^{2}−z^{2})`|(1,zx-y^2,xy-z^2),(0,(y-z),(z-x)),(0,(y-x),(z-y))|`

⇒∆=(x+y+z)^{2}(xy+yz+zx−x^{2}−y^{2}−z^{2})[{(y−z)(z−y)−(z−x)(y−x)}−0+0]

⇒∆=(x+y+z)^{2}(xy+yz+zx−x^{2}−y^{2}−z^{2})

So, ∆ is divisible by (*x* + *y* + *z*).

The quotient when ∆ is divisible by (*x* + *y* + *z*) is (x+y+z)(xy+yz+zx−x^{2}−y^{2}−z^{2})