Q. 105.0( 1 Vote )

Prove the followi

Answer :


Applying, C2C2 – 2C1 – 2C3, we get,




Taking, – (a2 + b2 + c2) common from C2 we get,



Applying R2R2 – R1 and R3R3 – R1, we get




= – (a2 + b2 + c2)(a – b)(c – a)[( – (b + a))( – b) – (c)(c + a)]


= (a – b)(b – c)(c – a)(a + b + c)(a2 + b2 + c2)


= R.H.S


Hence, proved.

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