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Question
Without expanding determinants, prove that `|(1, yz, y + z),(1, zx, z + x),(1, xy, x + y)| = |(1, x, x^2),(1, y, y^2),(1, z, z^2)|`.
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Solution
L.H.S. = `|(1, yz, y + z),(1, zx, z + x),(1, xy, x + y)|`
= `|(1, yz, x + y + z - x),(1, zx, y + z + x - y),(1, xy, z + x + y - z)|`
= `|(1, yz, x + y + z),(1, zx, x + y + z),(1, xy, x + y + z)| - |(1, yz, x),(1, zx, y),(1, xy, z)|`
= `x + y + z |(1, yz, 1),(1, zx, 1),(1, xy, 1)| - |(1, yz, x),(1, zx, y),(1, xy, z)|`
= `(x + y + z)0 - |(1, yz, x),(1, zx, y),(1, xy, z)|`
= `0 -|(1/x xx x, 1/x xx yz, 1/x xx x xx x),(1/yxxy, 1/yxxyxxzx, 1/yxxyxxy),(1/zxxz, 1/zxxzxx xy, 1/zxxzxxz)|`
Taking `1/x, 1/y and 1/z` from R1, R2 & R3.
`-1/(xyz)|(x, xyz, x^2),(y, xyz, y^2),(z, xyz, z^2)|`
Taking xyz from C2
`-(xyz)/(xyz)|(x, 1, x^2),(y, 1, y^2),(z, 1, z^2)|=-|(x,1,x^2),(y,1,y^2),(z,1,z^2)|`
C1 ↔ C2
= `|(1,x,x^2),(1,y,y^2),(1,z,z^2)|`
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