If \(a,b,c\) are real numbers and \(a+b+c\)=M , \(a^{3}+b^{3}+c^{3}=M^{3}\) \[\sum_{cyc}^{a,b,c}\frac{a^{3}}{1-a^{2}}=\frac{1}{6}\] . Then find M.
Can you please help me to solve this.

By given two conditions we get 3(a+b)(b+c)(c+a)=0
hence we can write a=-b and our summation reduces to c^3/1-c^2
Hence we find one of roots of this cubic as 0.5.
Hence M =0.5

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TopNewestBy given two conditions we get 3(a+b)(b+c)(c+a)=0

hence we can write a=-b and our summation reduces to c^3/1-c^2

Hence we find one of roots of this cubic as 0.5.

Hence M =0.5

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Can you please elaborate?

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\((a + b + c)^3 = a^3 + b^3 + c^3 + 3(a + b)(b + c)(c + a)\)a + b + c = M

\(a^3 + b^3 + c^3 = M^3\)

which is a = b = c = 0

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also a + b = 0

b + c = 0

c + a = 0

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