If $\frac{1}{x^{b} + x^{- c} + 1} + \frac{1}{x^{c} + x^{- a} + 1} + \frac{1}{x^{a} + x^{- b} + 1} = 1$ $1/(x^b+x^(-c)+1) + 1/(x^c+x^-a+1) + 1/(x^a+x^(-b)+1) = 1$ then what can we say about $a, b, c$ $a, b, c$ ?

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If $\frac{1}{x^{b} + x^{- c} + 1} + \frac{1}{x^{c} + x^{- a} + 1} + \frac{1}{x^{a} + x^{- b} + 1} = 1$ $1/(x^b+x^(-c)+1) + 1/(x^c+x^-a+1) + 1/(x^a+x^(-b)+1) = 1$ then what can we say about $a, b, c$ $a, b, c$ ?

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Answer 1 · 2016-08-21T08:37:34

$a + b + c = 0$ $a+b+c=0$

Explanation:

For any non-zero value of $x$ $x$ , we have $x^{0} = 1$ $x^0 = 1$ .

So with $a = b = c = 0$ $a=b=c=0$ we have:

$\frac{1}{x^{b} + x^{- c} + 1} + \frac{1}{x^{c} + x^{- a} + 1} + \frac{1}{x^{a} + x^{- b} + 1}$ $1/(x^b+x^(-c)+1) + 1/(x^c+x^-a+1) + 1/(x^a+x^(-b)+1)$

$= \frac{1}{3} + \frac{1}{3} + \frac{1}{3} = 1$ $=1/3+1/3+1/3 = 1$

Actually as seen in https://socratic.org/s/axdYQgwe, if $a + b + c = 0$ $a+b+c=0$ then this equation holds for any non-zero value of $x$ $x$ .

Note also that if $a = b = c = k$ $a=b=c=k$ then:

$1 = \frac{1}{x^{k} + x^{- k} + 1} + \frac{1}{x^{k} + x^{- k} + 1} + \frac{1}{x^{k} + x^{- k} + 1}$ $1 = 1/(x^k+x^(-k)+1) + 1/(x^k+x^-k+1) + 1/(x^k+x^(-k)+1)$

$= \frac{3}{x^{k} + x^{- k} + 1}$ $=3/(x^k+x^(-k)+1)$

So we have:

$x^{k} + x^{- k} + 1 = 3$ $x^k+x^(-k)+1 = 3$

Subtracting $3$ $3$ from both sides and multiplying through by $x^{k}$ $x^k$ we get:

$0 = {(x^{k})}^{2} - 2 (x^{k}) + 1 = {(x^{k} - 1)}^{2}$ $0 = (x^k)^2-2(x^k)+1 = (x^k-1)^2$

So $x^{k} = 1$ $x^k = 1$

This is satisfied for any non-zero value of $x$ $x$ if $k = 0$ $k = 0$ and no other values of $k$ $k$ .

Answer 2 · 2016-08-21T10:52:22

$a + b + c = 0$ $a+b+c=0$

Explanation:

Using "brute force" or with the help of a symbolic processor,

$\frac{1}{x^{b} + x^{- c} + 1} + \frac{1}{x^{c} + x^{- a} + 1} + \frac{1}{x^{a} + x^{- b} + 1} = \frac{n}{d} = 1$ $1/(x^b + x^-c + 1) + 1/(x^c + x^-a + 1) + 1/(x^a + x^-b + 1) =n/d=1$

$n = (x^a + x^b + 2 x^(a + b) + x^(2 a + b) + x^c + 2 x^(a + c) + 2 x^(b + c) + 6 x^(a + b + c) + 2 x^(2 a + b + c) + x^(2 b + c) + 2 x^(a + 2 b + c) + x^(2 a + 2 b + c) + x^(a + 2 c) + 2 x^(a + b + 2 c) + x^(2 a + b + 2 c) + x^(a + 2 b + 2 c))$

$d = (1 + x^a + x^b + 2 x^(a + b) + x^(2 a + b) + x^c + 2 x^(a + c) + 2 x^(b + c) + 4 x^(a + b + c) + 2 x^(2 a + b + c) + x^(2 b + c) + 2 x^(a + 2 b + c) + x^(2 a + 2 b + c) + x^(a + 2 c) + 2 x^(a + b + 2 c) + x^(2 a + b + 2 c) + x^(a + 2 b + 2 c) + x^( 2 a + 2 b + 2 c))$

but

$n-d = -1 + 2 x^(a + b + c) - x^(2 (a + b + c))$ so if $n = d$

$-1 + 2 x^(a + b + c) - x^(2 (a + b + c))=0$

Solving for $x^(a + b + c)$ we obtain

$x^(a + b + c) = 1$ then $a+b+c=0$

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If $\frac{1}{x^{b} + x^{- c} + 1} + \frac{1}{x^{c} + x^{- a} + 1} + \frac{1}{x^{a} + x^{- b} + 1} = 1$ $1/(x^b+x^(-c)+1) + 1/(x^c+x^-a+1) + 1/(x^a+x^(-b)+1) = 1$ then what can we say about $a, b, c$ $a, b, c$ ?

2 Answers

Explanation:

Explanation:

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If 1/(x^b+x^(-c)+1) + 1/(x^c+x^-a+1) + 1/(x^a+x^(-b)+1) = 11xb+x−c+1+1xc+x−a+1+1xa+x−b+1=11/(x^b+x^(-c)+1) + 1/(x^c+x^-a+1) + 1/(x^a+x^(-b)+1) = 1 then what can we say about a, b, ca,b,ca, b, c ?

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Explanation:

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If $\frac{1}{x^{b} + x^{- c} + 1} + \frac{1}{x^{c} + x^{- a} + 1} + \frac{1}{x^{a} + x^{- b} + 1} = 1$ $1/(x^b+x^(-c)+1) + 1/(x^c+x^-a+1) + 1/(x^a+x^(-b)+1) = 1$ then what can we say about $a, b, c$ $a, b, c$ ?