If a,b,c are all distinct and `|[a,a^3,a^4-1],[b,b^3,b^4-1],[c,c^3,c^4-1]|` =0, show that abc(ab+bc+ac) = a+b+c

Explanation

we have `|{:(a,,a^(3),,a^(4)),(b,,b^(3),,b^(4)),(c,,c^(3),,c^(4)):}|-|{:(a,,a^(3),,1),(b,,b^(3),,1),(c,,c^(3),,1):}|=0 "(Splitting w.r.t " C_(3)")"`
`:. Abc |{:(1,,a^(2),,a^(3)),(1,,b^(2),,b^(3)),(1,,c^(2),,c^(3)):}|-|{:(1,,a,,a^(3)),(1,,b,,b^(3)),(1,,c,,c^(3)):}|=0`
`rArr abc(a -b) (b-c) (c-a) (ab+bc+ca)`
`-(a-b)(b-c) (c-a)(a+b+c) =0`
`rArr abc (ab+bc+ca) -(a+b+c)=0`
`(As a,b,c " are distinct)"`