Question

solve fast.........pls
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Answer 1

Answer:

Given:

$\sf{a+b+c=0}$

Solution:

$\sf{a+b+c=0}$

$\sf{\therefore{-c=a+b \ and \ c=-a-b}}$

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

$\sf{Multiply \ first \ term \ by \ x^{b}}$

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

$\sf{In \ second \ term \ by \ substitute \ -c=a+b}$

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

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

$\sf{Substitute \ c=-a-b \ in \ third \ term}$

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

$\sf{Multiply \ third \ term \ by \ x^{a+b}}$

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

$\sf{\leadsto{\dfrac{x^{b}}{x^{a+b}+x^{b}+1}+\dfrac{1}{x^{a+b}+x^{b}+1}+\dfrac{x^{a+b}}{x^{a+b}+x^{b}+1}}}$

$\sf{\leadsto{\dfrac{x^{b}+1+x^{a+b}}{x^{a+b}+x^{b}+1}}}$

$\sf{\leadsto{1}}$

$\sf\purple{\tt{Hence, \ the \ answer \ is \ 1.}}$

Answer 2

aapka naam...?...???