Q. 215.0( 3 Votes )

If a + b + c ≠ 0

Answer :

Let

Applying C1 C1 + C2 + C3, we get



Taking (a + b + c) common from the first column, we get



Now, Expanding along C1, we get


= (a + b + c)[(1)(bc – a2) – (1)(b2 – ac) + (1)(ba – c2)]


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


= (a + b + c)[-(a2 + b2 + c2 – ab – bc – ac)]





[ (a – b)2 = a2 + b2 – 2ab]


Given that Δ = 0



(a + b + c)[(a – b)2 + (b – c)2 + (c – a)2] = 0


Either (a + b + c) = 0 or (a – b)2 + (b – c)2 + (c – a)2 = 0


but it is given that (a + b + c) ≠ 0


(a – b)2 + (b – c)2 + (c – a)2 = 0


a – b = b – c = c – a = 0


a = b = c


Hence Proved


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