If `|{:(a,,a^(2),,1+a^(3)),(b,,b^(2),,1+b^(3)),(c,,c^(2),,1+c^(3)):}|=0` and the vectors
`vecA =(1, a, a^(2)) , vec(B) = (1, b, b^(2)) , vec(C )(1,c,c^(2))`
are non-coplanar then the product abc = ….

Answer

`-1`

Explanation

Given that `|{:(a,, a^(2),, 1+a^(3)), (b,,b^(2),,1+b^(3)),(c,,c^(2),,1+c^(3)):}|=0`
`|{:(a,,a^(2),,1),(b,,b^(2),,1),(c,, c^(2),,1):}|+ abc|{:(1,,a,,a^(2)),(1,,b,,b^(2)),(1,,c,,c^(2)):}|=0`
Operating `C_2 harr C_3 and ` then `C_1 harr C_2` in first determinant
`|{:(1,,a,,a^(2)),(1,,b,,b^(2)),(1,,c,,c^(2)):}|+abc |{:(1,,a,,a^(2)),(1,,b,,b^(2)),(1,,c,,c^(2)):}| =0`
either 1 + abc =0 or `|{:(1,,a ,,a^(2)),(1,,b,,b^(2)),(1,,c,,c^(2)):}|=0`
Also given that vectors `vecA, vecB and vecC` are non-coplanar.
`|{:(1,,a,,a^(2)),(1,,b,,b^(2)),(1,,c,,c^(2)):}| ne 0`
So we must have `1+ abc =0 or abc =-1`