Answer :

Given,

Performing the operation C_{1}→ C_{1} – C_{2} and C_{2}→ C_{2} – C_{3}, we get

__We know that a ^{3} – b^{3} = (a – b) (a^{2} + ab + b^{2})__

Taking (a – b) and (b – c) common from C_{1} and C_{2} respectively, we get

Expanding along R_{1}, we get

= (a – b) (b – c) [(b^{2} + bc + c^{2}) – (a^{2} + ab + b^{2})]

= (a – b) (b – c) [(c^{2} – a^{2}) + (bc – ab)]

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

= (a – b) (b – c) (c – a) (c + a + b)

= (a – b) (b – c) (c – a) (a + b + c)

= RHS

Hence proved.

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