Question

could you please solve this?​

Answer 1

Answer:

$=\dfrac{1}{1+x^a+x^{-b}}+\dfrac{1}{1+x^b+x^{-c}}+\dfrac{1}{1+x^c+x^{-a}}$

$=\dfrac{x^b}{x^b(1+x^a+x^{-b})}+\dfrac{1}{1+x^b+x^{a+b}}+\dfrac{1}{1+x^c+x^{-a}}$

$=\dfrac{x^b}{x^b+x^{a+b}+1}+\dfrac{1}{1+x^b+x^{a+b}}+\dfrac{1}{1+x^c+x^{-a}}$

$=\dfrac{x^b+1}{x^b+x^{a+b}+1}+\dfrac{1}{1+x^c+x^{-a}}$

$=\dfrac{x^b+1}{x^b+x^{a+b}+1}+\dfrac{x^{a+b}}{x^{a+b}(1+x^c+x^{-a})}$

$=\dfrac{x^b+1}{x^b+x^{a+b}+1}+\dfrac{x^{a+b}}{x^{a+b}+x^{c+a+b}+x^{-a+a+b}}$

$=\dfrac{x^b+1}{x^b+x^{a+b}+1}+\dfrac{x^{a+b}}{x^{a+b}+x^{0}+x^{b}}$

$=\dfrac{x^b+1}{x^b+x^{a+b}+1}+\dfrac{x^{a+b}}{x^{a+b}+1+x^{b}}\\\\=\dfrac{x^b+1+x^{a+b}}{x^b+x^{a+b}+1}\\\\=1$

Answer 2

Answer:

Proof is provided below

Step-by-step explanation:

If a + b + c = 0

We will be making use of these throughout the proof

b + c = - a

a + b = - c

a + c = - b

Prove
$\dfrac{1}{1 + x^a + x^{-b}}+ \dfrac{1}{1 + x^b + x^{-c}}+\dfrac{1}{1 + x^c + x^{-a}} = 1$

Take the first expression:
$\dfrac{1}{1 + x^a + x^{-b}} = \dfrac{1}{1 + x^a + \dfrac{1}{x^{b}}} \\\\=\dfrac{ \dfrac{1}{x^b + x^ax^b + 1}}{x^b}\\\\= \dfrac{x^b}{1 + x^{a+b}+x^b}\\\\= \dfrac{x^b}{1 + x^{-c}+x^b}\quad(a+ b = -c)\\\\\\= \dfrac{x^b}{1 + x^{b}+x^{-c}} \quad \text{(Re-arranging terms in denominator})$

The original expression now becomes

$\dfrac{x^b}{1 + x^{b}+x^{-c}} + \dfrac{1}{1 + x^b + x^{-c}}+\dfrac{1}{1 + x^c + x^{-a}}$

The first two terms have the same denominator so we can add:
$\dfrac{x^b+1}{1 + x^b + x^{-c}}+\dfrac{1}{1 + x^c + x^{-a}}$

Simplify the first term as follows:

$\dfrac{x^b+1}{1 + x^b + x^{-c}}\\\\= \dfrac{x^b+1}{1 + x^b + \dfrac{1}{x^c}}\\\\\\= \dfrac{x^b+1}{ \dfrac{x^c + x^bx^c + 1}{x^c}}\\\\\\= \dfrac{x^c(x^b + 1)}{x^c+ x^{b+c}+ 1}\\\\= \dfrac{x^{b+c} + x^c}{x^c + x^{-a} + 1} \quad\text(b + c = -a)}$

Therefore:
$\dfrac{x^b+1}{1 + x^b + x^{-c}}+\dfrac{1}{1 + x^c + x^{-a}} \\\\= \dfrac{x^{b+c} + x^c}{x^c + x^{-a} + 1} + \dfrac{1}{1 + x^c + x^{-a}}$

Denominator same:
$= \dfrac{x^{b+c} + x^c+1}{1 + x^c + x^{-a}}$

= $\dfrac{x^{-a} + x^c + 1}{1 + x^c + x^{-a}}$

Rearranging the numerator we get
$\dfrac{1+ x^{c} + x^{-a} }{1 + x^c + x^{-a}}\\\\= 1$