Question

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.

Answer 1

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 C₁→C₁+C₂+C₃ , 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−x²−y²−z²)`|(1,zx-y^2,xy-z^2),(1,xy-z^2,yz-x^2),(1,yz-x^2,zx-y^2)|`

Applying R₂→R₂−R₁ and R₃→R₃−R₁, we get

∆=(xy+yz+zx−x²−y²−z²)`|(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−x²−y²−z²)`|(1,zx-y^2,xy-z^2),(0,(y-z),(z-x)),(0,(y-x),(z-y))|`

⇒∆=(x+y+z)²(xy+yz+zx−x²−y²−z²)[{(y−z)(z−y)−(z−x)(y−x)}−0+0]

⇒∆=(x+y+z)²(xy+yz+zx−x²−y²−z²)

So, ∆ is divisible by (x + y + z).

The quotient when ∆ is divisible by (x + y + z) is (x+y+z)(xy+yz+zx−x²−y²−z²)