Using properties of Determinants, prove that:-
= 2(a+b)(b+c)(c+a)

Let  = a+b+c-c-b-ca+b+c-a-b-aa+b+cApplying R1 = R1 + R3 = a+c-a+ca+c-ca+b+c-a-b-aa+b+c = a+c1-11-ca+b+c-a-b-aa+b+capplying C1 = C1 + C2  and C2 = C2 + C3 =a+c001a+bb+c-a-a+bb+ca+b+c = a+bb+cc+a00111-a-11a+b+c = a+bb+cc+a 11+1 = 2a+bb+cc+a

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First do C1+C2, you get (a+b) common in 1st column .
Then do C2+C3, you get (b+c) common in 2nd column.
Then solve the determinant.:):)
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