Finding $\frac {8}{x^{a-b}+1}+\frac {8}{x^{b-a}+1} = ?$

Question

$$\frac {8}{x^{a-b}+1}+\frac {8}{x^{b-a}+1} = ?$$

My attempt:

$$\frac {8}{x^{a-b} + 1} = 8. x^{-a+b}-1 \tag 1$$

$$\frac {8}{x^{b-a} + 1} = 8. x^{a-b}-1 \tag 2$$

I think I've gone too wrong.

Answer 1

$\cfrac 8{x^{b-a}+1}=\cfrac8{\frac{x^{b}}{x^a}+1}=\cfrac {8x^a}{x^b+x^a}$ after multiplying both numerator and denominator by $x^a$

$\cfrac 8{x^{a-b}+1}=\cfrac8{\frac{x^{a}}{x^b}+1}=\cfrac {8x^b}{x^a+x^b}$ after multiplying both numerator and denominator by $x^b$

$$\cfrac {8x^b}{x^a+x^b}+\cfrac {8x^a}{x^b+x^a}=?$$

Answer 2

Let $a-b=t$ for sometime. You must be knowing that $$x^{-t}=\frac1{x^t}$$

Let $x^t=y$

Therefore you have \begin{align} \frac{8}{x^{a-b}}+\frac{8}{x^{b-a}+1} &=\frac{8}{y+1}+\frac{8}{1/y+1}\\ &= \frac{8}{y+1} + \frac{8y}{1+y}\\ &=\frac{8y+8}{y+1} \\&= \frac{8(y+1)}{y+1} \\&=8 \end{align}

Answer 3

$$\frac {8}{x^{a-b}+1}+\frac {8}{x^{b-a}+1} = \frac {8}{x^{a-b}+1}+\frac {8}{\frac{1}{x^{a-b}}+1}=\frac {8}{x^{a-b}+1}+\frac {8x^{a-b}}{x^{a-b}+1}=8.$$

Answer 4

let $n=a-b$ \begin{eqnarray*} \frac{8}{x^n+1}+ \frac{8}{x^{-n}+1} =\frac{8}{x^n+1}+ \frac{8x^n}{x^{n}+1} =\frac{8(1+x^n)}{1+x^n}=\color{red}{8}. \end{eqnarray*}