Question

Simplify for $A$ and $B$ and find their sum, if $A=\frac{1}{1+x^{b-a}+x^{c-a}}+\frac{1}{1+x^{c-b}+x^{a-b}}+\frac{1}{1+x^{a-c}+x^{b-c}}$ and
$B={\left(\frac{x^m}{x^n}\right)}^{1/(m-n)}\times {\left(\frac{x^n}{x^l}\right)}^{1/(n-l)}\times {\left(\frac{x^l}{x^m}\right)}^{1/(l-m)}$ then A+B

Answer 1

We have,

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

$A=\frac{1}{1+x^b/x^a+x^c/x^a}+\frac{1}{1+x^c/x^b+x^a/x^b}+\frac{1}{1+x^a/x^c+x^b/x^c}$

$A=\frac{x^a}{x^a+x^b+x^c}+\frac{x^b}{x^b+x^c+x^a}+\frac{x^c}{x^c+x^a+x^b}$

$A=\frac{x^a+x^b+x^c}{x^a+x^b+x^c}=1$

And

$B={\left(\frac{x^m}{x^n}\right)}^{1/(m-n)}\times {\left(\frac{x^n}{x^l}\right)}^{1/(n-l)}\times {\left(\frac{x^l}{x^m}\right)}^{1/(l-m)}$

$B={\left(x^{m-n}\right)}^{1/(m-n)}\times {\left(x^{n-l}\right)}^{1/(n-l)}\times {\left(x^{l-m}\right)}^{1/(l-m)}$

$B=1\times 1\times 1=1$

Therefore,

$A+B=1+1=2$

Hence, this is the answer.