Answer :

Let Δ =

Applying Elementary Transformations

C_{1}→ C_{1} + C_{2} + C_{3}, we have

Δ =

Taking (a + b + c) common from C_{1}, we get

Δ = (a + b + c)

Applying R_{2}→ R_{2} – R_{1} and R_{3}→ R_{3} – R_{1}

Expanding along C_{1}

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

Δ = (a + b + c) [4bc + 2ab + 2ac + a^{2} – a^{2} + 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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