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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.
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 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)
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