Using properties of determinants, prove the following: \begin{vmatrix} 3a&-a+b&-a+c\\ a-b&3b&c-b\\ a-c&b-c&3c \end{vmatrix} = `3(a+b+c)(ab+bc+ca)`

Explanation

`L.H.S. = |[3a,-a+b,-a+c],[a-b,3b,c-a],[a-c,b-c,3c]|`

Applying `C_1->C_1+C_2+C_3`

`=|[a+b+c,-a+b,-a+c],[a+b+c,3b,c-a],[a+b+c,b-c,3c]|`

`=(a+b+c)|[1,-a+b,-a+c],[1,3b,c-a],[1,b-c,3c]|`

Applying `R_2->R_2-R_1 and R_3->R_3-R_1`

`=(a+b+c)|[1,-a+b,-a+c],[0,2b+a,a-b],[0,a-c,2c+a]|`

`=(a+b+c)[(2b+a)(2c+a) - (a-b)(a-c)]`

`=(a+b+c)[4bc+2ab+2ac+a^2 - (a^2+bc-ab-ac)]`

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