prove that `|{:((b+c)^(2),,bc,,ac),(ba,,(c+a)^(2),,cb),(ca,,cb,,(a+b)^(2)):}|`
`|{:((b+c)^(2),,a^(2),,a^(2)),(b^(2),,(c+a)^(2),,b^(2)),(c^(2),,c^(2),,(a+b)^(2)):}| =2abc (a+b+c)^(3)`

Explanation

Multiplying `R_(1)R_(2),R_(3)` by a,b,c respectively and dividing by abc we get `Delta =(1)/(abc)xx |{:(a(b+c)^(2),,ba^(2),,ca^(2)),(ab^(2),,b(c+a)^(2),,cb^(2)),(ac^(2),,bc^(2),,c(a+b)^(2)):}|`
Taking a,b,c common from` C_(1),C_(2) " and " C_(3) ` respectively we get
`Delta = |{:((b+c)^(2),,a^(2),,a^(2)),(b^(2),,(c+a)^(2),,b^(2)),(c^(2),,c^(2),,(a+b)^(2)):}|`
` " Now applying " C_(2) to C_(2) -C_(1),C_(3) to C_(3) -C_(1)` gives
`Delta =|{:((b+c)^(2),,(a+b+c)(a-b-c),,(a-b-c)(a+b+c)),(b^(2),,(c+a+c)(c+a-b),,0),(c^(2),,0,,(a+b+c)(a+b-c)):}|`
` (a+b+c)^(2) xx |{:((b+c)^(2),,a-b-c,,a-b-c),(b^(2),,c+a-b,,0),(c^(2),,0,,a+b-c):}|`
Applying `R_(1) to R_(1) -(R_(2)+R_(3))` and then taking 2 common from `R_(1)` we get
`Delta =2(a+b+c)^(2) xx |{:(bc,,-c,,-b),(b^(2),,c+a-b,,0),(c^(2),,0,,a+b-c):}|`
`" Now applying "C_(2) to bC_(2) +C_(1), C_(3) to cC_(3) +C_(1)` gives
`Delta =(2(a+b+c)^(2))/(bc) xx |{:(bc,,0,,0),(b^(2),,b(c+a),,b^(2)),(c^(2),,c^(2),,c(a+b)):}|`
`=(2(a+b+c)^(2))/(bc) bc[(bc +ba)(ca+cb)-b^(2)c^(2)]`
`=2(a+b+c)^(2) [bc(ac+bc+ab+a^(2)-bc)]`
`2abc (a+b+c)^(3)`