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a
(z−x)(y−z)(y−x+xyz)
b
(x−z)(z−y)(y−x+xyz)
c
(z+x)(y+z)(y+x−xyz)
d
(z+x)(y−z)(y+x−xyz)
answer is A.
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Detailed Solution
We have , 0 xyz x-z y-x 0 y-zz-x z - y 0=z-x xyz x-z z-x 0 y-zz-x z - y 0 ∵ C1→C1- C3 =(z-x)1 xyz x-z 1 0 y-z1 z - y 0 taking(z-x)common from column 1 (Expanding along R1) =(z-x)[1.{-(y-z)(z-y)-xyz(z-y)+(x-z)(z-y)]=(z-x)(z-y)(-y+z-xyz+x-z)=(z-x)(z-y)(x-y-xyz)=(z-x)(y-z)(y-x+xyz)