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If 1/a+1/b+1/c =0then a^2+b^2+c^2= |
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Answer» now, 1/a²+b²+c² + 1/b²+c²-a² + 1/c²+a²-b² =1/a²+(b+c)(b-c) + 1/b²+(c+a)(c-a) + 1/c²+(c+b)(a-b) =1/a²+(-a)(b-c) + 1/b²+ (-b)(c-a) + 1/c²+ (-c)(a-b) (from 1) =1/a(a-b+c) + 1/b(a+b-c) = 1/c(c-a+b) =1/a(a+c-b) + 1/b(a+b-c) + 1/c(b+c-a) =1/a(-b-b) + 1/b(-c-c) + 1/ c (-a-a) (from 1) =1/-2ab + 1/-2bc + 1/-2ac = 1/-2 (1/ab + 1/bc+ 1/ac ) = -1/2 (a+b+c/abc) =-1/2 (0) (from 1) = 0 |
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