Answer :
Applying, C2→C2 – 2C1 – 2C3, we get,
Taking, – (a2 + b2 + c2) common from C2 we get,
Applying R2→R2 – R1 and R3→R3 – 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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