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Show that if x(1),x(2),x(3) ne 0 |{:(x...

Show that if `x_(1),x_(2),x_(3) ne 0`
`|{:(x_(1) +a_(1)b_(1),,a_(1)b_(2),,a_(1)b_(3)),(a_(2)b_(1),,x_(2)+a_(2)b_(2),,a_(2)b_(3)),(a_(3)b_(1),,a_(3)b_(2),,x_(3)+a_(3)b_(3)):}|`
`=x_(1)x_(2)x_(3) (1+(a_(1)b_(1))/(x_(1))+(a_(2)b_(2))/(x_(2))+(a_(3)b_(3))/(x_(3)))`

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The given determinant can be written as the sum of two determinants
`|{:(x_(1),,a_(1)b_(2),,a_(1)b_(3)),(0,,x_(2)+a_(2)b_(2),,a_(2)b_(3)),(0,,a_(3)b_(2),,x_(3)+a_(3)b_(3)):}|+|{:(a_(1)b_(1),,a_(1)b_(2),,a_(1)b_(3)),(a_(2)b_(1),,x_(2)+a_(2)b_(2),,a_(2)b_(3)),(a_(3)b_(1),,a_(3)b_(2),,x_(3)+a_(3)b_(3)):}|`
Expanding the first determinant along `C_(1)` we get
`x_(1) |{:(x_(2)+a_(2)b_(2),,a_(2)b_(3)),(a_(3)b_(2),,x_(3)+a_(3)b_(3)):}|`
`=x_(1)[(x_(2)+a_(2)b_(2))(x_(3)+a_(3)b_(3))-a_(3)b_(2)a_(2)b_(3)]`
`=x_(1)(x_(2)x_(3)+x_(3)a_(2)b_(2)+x_(2)a_(3)b_(3)+a_(2)b_(2)a_(3)b_(3)-a_(3)b_(2)b_(3)]`
`=x_(1)x_(2)x_(3)+x_(1)x_(3)a_(2)b_(2)+x_(1)x_(2)a_(3)b_(3)`
In the second determinant taking `b_(1)` common from `C_(1)` and then applying `C_(1) to C_(2) -b_(2)C_(1)" and " C_(3) to -b_(3)dC_(3)` we obtain
`b_(1) |{:(a_(1),,0,,0),(a_(2),,x_(2),,0),(a_(3),,0,,x_(3)):}|=a_(2)b_(1)x_(2)x_(3)`
Therefore the given determinant is
`x_(1)x_(2)x_(3)+x_(1)x_(3)a_(2)b_(2)+x_(1)x_(2)a_(3)b_(3)+a_(1)b_(1)x_(2)x_(3)`
`=x_(1)x_(2)x_(3) (1+(a_(1)b_(1))/(x_(1)) +(a_(2)b_(2))/(x_(2)) +(a_(3)b_(3))/(x_(3)))`
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