Using properties of determinants, prove that

To Prove:

Proof:

Applying R₁→ R₁ + R₃ and R₂→ R₂ + R₃

Now, taking (a + c) common from the first row and (b + c) common from the second row, we have,

Expanding, we get,

= (a + c)(b + c)[a + b + c + a + (- a – b – c + b) + 1(a + b)]

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

= 2(a + b)(b + c)(c + a)

= R.H.S

Hence, Proved.