# Doubt on solving equations

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.

2 years, 11 months ago

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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

- 2 years, 11 months ago

- 2 years, 10 months ago

$$(a + b + c)^3 = a^3 + b^3 + c^3 + 3(a + b)(b + c)(c + a)$$

a + b + c = M

$$a^3 + b^3 + c^3 = M^3$$

which is a = b = c = 0

- 2 years, 10 months ago

also a + b = 0

b + c = 0

c + a = 0

- 2 years, 10 months ago