×

# 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, 4 months ago

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold
- bulleted- list
• bulleted
• list
1. numbered2. list
1. numbered
2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> This is a quote
This is a quote
    # I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears 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} $$\boxed{123}$$

Sort by:

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, 4 months ago

- 2 years, 4 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, 3 months ago

also a + b = 0

b + c = 0

c + a = 0

- 2 years, 3 months ago