Answer :

Let Δ =

Applying Elementary Transformations


C1 C1 + C2 + C3, we have


Δ =


Taking (a + b + c) common from C1, we get


Δ = (a + b + c)


Applying R2 R2 – R1 and R3 R3 – R1



Expanding along C1


Δ = (a + b + c) [1 × {(2b + a) (2c + a) – (a – b) (a – c)} – 0 + 0]


Δ = (a + b + c) [4bc + 2ab + 2ac + a2 – a2 + ac + ba – bc]


Δ = (a + b + c) (3ab + 3bc + 3ac)


Δ = 3 (a + b + c) (ab + bc + ca)


Hence, the given result proved.


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