if `Delta (x)= |{:(a_(1)+x,,b_(1)+x,,c_(1)+x),(a_(2)+x,,b_(2)+x,,c_(2)+x),(a_(3)+x,,b_(3)+x,,c_(3)+x):}|` then show that `Delta (x)=0`
and that `Delta (x)=Delta(0)+sx.` where s denotes the sum of all the cofactors of all the elements in `Delta (0)`

Explanation

`Delta (x)= |{:(a_(1)+x,,b_(1)+x,,c_(1)+x),(a_(2)+x,,b_(2)+x,,c_(2)+x),(a_(3)+x,,b_(3)+x,,c_(3)+x):}|`
` :. Delta (x) = |{:(a_(1)+x,,b_(1)+x,,c_(1)+x),(a_(2)+x,,b_(2)+x,,c_(2)+x),(a_(3)+x,,b_(3)+x,,c_(3)+x):}|+|{:(a_(1)+x,,1,,c_(1)+x),(a_(2)+x,,1,,c_(2)+x),(a_(3)+x,,1,,c_(3)+x):}|`
`+|{:(a_(1)+x,,b_(1)+x,,1),(a_(2)+x,,b_(2)+x,,1),(a_(3)+x,,b_(3)+x,,1):}|`
Applying `C_(2) to C_(2)-xC_(1),C_(3) to C_(3) xC_(1)` in the first det.
`C_(1) to C_(1) -xC_(2),C_(3) to C_(3) -xC_(3)-xC_(2)` in the second det.
`" and " C_(1) to C_(1)-xC_(3),C_(2)to C_(2)-xC_(3)` in the third det. we get
`Delta (x)= |{:(1,,b_(1),,C_(1)),(1,,b_(2),,c_(2)),(1,,b_(3),,c_(3)):}|+|{:(a_(1),,1,,c_(1)),(a_(2),,1,,c_(2)),(a_(3),,1,,c_(3)):}|+|{:(a_(1),,b_(1),,1),(a_(2),,b_(2),,1),(a_(3),,b_(3),,1):}|`
`Delta (0)= |{:(a_(1),,b_(1),,c_(1)),(a_(2),,b_(2),,c_(2)),(a_(3),,b_(3),,c_(3)):}|`
which are `b_(2) C_(3) =b_(3)C_(2),C_(2)a_(3)-C_(3)a_(2),a_(2)b_(3)-b_(2)a_(3)` etc
Clearly
` |{:(1,,b_(1),,c_(1)),(1,,b_(2),,c_(2)),(1,,b_(3),,c_(3)):}| = (b_(2)b_(3) -b_(3)c_(2))+(c_(1)b_(3)-c_(3)b_(1))+(b_(1)c_(2)-b_(2)c_(1))`
Which is the sum of cofactors of the first row elements of `Delta (0)`
`" similarly " |{:(a_(1),,1,,c_(1)),(a_(2),,1,,c_(2)),(a_(3),,1,,c_(3)):}|" and " |{:(a_(1),,b_(1),,1),(a_(2),,b_(2),,1),(a_(3),,b_(3),,1):}|` are the sum of
cofactors of 2nd row and 3rd elements respectively of `Delta (0)` .
Hence `Delta(x)=s` where S denotes the sum of all cofactors of elements of `Delta(0)`
`:. Delta ''(x) =0`
since `Delta (x) = s Delta (x) sx+k`
So `Delta (0) =k`
Hence `Delta(x)= x S+Delta (0)`