Prove the identities: `|b^2+c^2a b a c b a c^2+a^2b cc a c b a^2+b^2|=4a^2b^2c^2`

Explanation

`Delta = |{:(b^(2)+c^(2),,ab,,ac),(ab,,c^(2)+a^(2),,bc),(ca,,ca,,a^(2)+b^(2)):}|`
Applying `R_(1) to aR_(1),R_(2) to bR_(2) " and " R_(3) to cR_(3)` we get
`Delta=(1)/(abc)xx |{:(a(b^(2)+c^(2)),,a^(2)b,,a^(2)c),(ab^(2),,b(c^(2)+a^(2)),,cb^(2)),(ac^(2),,bc^(2),,c(a^(2)+b^(2))):}|`
Taking a,b and c common from `C_(1),C_(2) " and "C_(3)` respectively we get
`Delta= |{:(b^(2)+c^(2),,a^(2),,a^(2)),(b^(2),,c^(2)+a^(2),,b^(2)),(c^(2),,c^(2),,a^(2)+b^(2)):}|`
Applying `R_(1) to R_(1)-R_(2)-R_(3)`
`Delta = |{:(0,,-2c^(2),,-2b^(2)),(b^(2),,c^(2)+a^(2),,b^(2)),(c^(2),,c^(2),,a^(2)+b^(2)):}|`
Taking 2 common from `R_(1) ` and applying `R_(2) toR_(2) +R_(1) " and "R_(3) to R_(3)+R_(1)`
`Delta= 2 |{:(0,,-c^(2),,-b^(2)),(b^(2),,a^(2),,0),(c^(2),,0,,a^(2)):}|`
Expanding along `R_(1)` we get
`Delta =2[c^(2)(a^(2)b^(2))-b^(2)(-a^(2)c^(2))]=4a^(2)b^(2)c^(2)`