If #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# ?

2 Answers
Aug 21, 2016

Answer:

#a+b+c=0#

Explanation:

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

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

#1/(x^b+x^(-c)+1) + 1/(x^c+x^-a+1) + 1/(x^a+x^(-b)+1)#

#=1/3+1/3+1/3 = 1#

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

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

#1 = 1/(x^k+x^(-k)+1) + 1/(x^k+x^-k+1) + 1/(x^k+x^(-k)+1)#

#=3/(x^k+x^(-k)+1)#

So we have:

#x^k+x^(-k)+1 = 3#

Subtracting #3# from both sides and multiplying through by #x^k# we get:

#0 = (x^k)^2-2(x^k)+1 = (x^k-1)^2#

So #x^k = 1#

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

Aug 21, 2016

Answer:

#a+b+c=0#

Explanation:

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

#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#