1.

The number of positive integral solutions of the equation `|(x^3+1,x^2y,x^2z),(xy^2,y^3+1,y^2z),(xz^2,z^2y,z^3+1)|=11` is

Answer» Correct Answer - B
`|{:(x^(3)+1,,x^(2)y,,x^(2)z),(xy^(2),,y^(3)+1,,y^(2)z),(xz^(2),,yz^(2),,z^(3)+1):}|=11`
Multiplying `R_(1) " by " x, R_(2)` by y and `R_(3) ` by z we bet
`(1)/(xyz) |{:(x^(4)+x,,x^(2)y,,x^(3)z),(xy^(3),,y^(4)+y,,y^(3)z),(xz^(3),,yz^(3),,z^(4)+z):}|=11`
Taking x,y,z common from `C_(1) ,C_(2),C_(3)` respectively we get
`|{:(x^(4)+x,,x^(2)y,,x^(3)z),(xy^(3),,y^(4)+y,,y^(3)z),(xz^(3),,yz^(3),,z^(4)+z):}|=11`
Using `R_(1) to R_(1) +R_(2)+R_(3)` we get
`(x^(3)+y^(3)+z^(3)+1) |{:(1,,1,,1),(y^(2),,y^(3)+1,,y^(3)),(z^(3),,z^(3),,z^(3)+1):}|=11`
Using `C_(2) to C_(2)-C_(1) " and " C_(3) toC_(3)-C_(1)` we get
`(x^(3) +y^(3) +z^(3) +1) |{:(1,,0,,0),(y^(3),,1,,0),(z^(2),,0,,1):}|=11`
Hence `,x^(3) +y^(3)+z^(3)=10`
Therefore the ordered triplets are (2,1,1) ,(1,2,1),(1,1,2)


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