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TopNewestThis was my solution : the a(b-c)^3 + b(c-a)^3 + c(a-b)^3 = 0 can be reduced to a(b-c)(b^2 + c^2) + b(c-a)(a^2 + c^2) + c(a-b)(b^2 + a^2) ...

Now there are 3 equations in terms of (b-c); (c-a); (a-b).

say b-c = x ; c-a = y ; a-b= z.

then we have x + y + z = 0;(self evident) a(x) + b(y) + c(z) = 0(self evident) (a(b^2+c^2))x + (b(a^2+c^2))y + (c(b^2+a^2))z = 0(given condition)

this is a system of equations x y and z .. {cij}[xi] = 0 .. trivial solution is x = y = z = 0;

for non trivial solutions determinant of {cij} = 0

the coefficients are in this form -- 1 1 1 a b c (a(b^2+c^2)) (b(a^2+c^2)) (c(b^2+a^2)) determinant = 0 gives us a condition a+b+c = 0; solving for this condition alone with the other given conditions that -10<=a<b<c gives us the required solutions..

Now what is wrong with this solution? – Raviteja Meesala · 4 years ago

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