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.

#Algebra #Summation #IITJEE #sequence #Algebricmanipulation

Easy Math Editor

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Comments

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

Aakash Khandelwal - 5 years, 9 months ago

Can you please elaborate?

Shivam Jadhav - 5 years, 9 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

Dev Sharma - 5 years, 9 months ago

also a + b = 0

b + c = 0

c + a = 0

Dev Sharma - 5 years, 9 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$