Using factor theorem, show that `|(-2a,a+b,c+a),(a+b,-2a,b+c),(c+a,c+b,-2c)|=4(a+b)(b+c)(c+a)`

Explanation

Let
`Delta =|{:(-2a,,a+b,,a+c),(b+a,,-2b,,b+c),(c+a,,c+b,,-2c):}|`
Putting `a+b =0 " or " b=-a ` we get
` Delta =|{:(-2a,,0,,a+c),(0,,2a,,c-a),(c+a,,c-a,,-2c):}|`
Expanding along `R_(1)`
`=-2a {-4ac -(c-a)^(2)}-0 +(a+c) {0 -2a(c+a)}`
`=2a(c+a)^(2) -2a(c+a)^(2)`
`=0`
Hence a+b is a factor of `Delta`
Similarly b+c and c+a are the factors of `Delta`
on expansion of determinant we can see that each terms of the determinaint is a homogeneous expression in a,b,c of degree 3 and also `R.H.S.` is a homogeneous expression of degree 3.
`|{:(-2a,,a+b,,a+c),(b+a,,-2b,,b+c),(c+a,,c+b,,-2c):}| =4 (a+b)(b+c) (c+a)`