The value of $$\frac{1}{{{a^2} + ax + {x^2}}}$$   $$ - $$ $$\frac{1}{{{a^2} - ax + {x^2}}}$$   $$ + $$ $$\frac{2ax}{{{a^4} + {a^2}{x^2} + {x^4}}}$$    is?

Solution(By Examveda Team)

$$\frac{1}{{{a^2} + ax + {x^2}}}$$   $$ - $$ $$\frac{1}{{{a^2} - ax + {x^2}}}$$   $$ + $$ $$\frac{2ax}{{{a^4} + {a^2}{x^2} + {x^4}}}$$
$$ = \frac{{{a^2} - ax + {x^2} - {a^2} - ax - {x^2}}}{{\left( {{a^2} + {x^2} + ax} \right)\left( {{a^2} + {x^2} - ax} \right)}} + $$       $$\frac{{2ax}}{{{a^4} + {a^2}{x^2} + {x^4}}}$$
$$\eqalign{ & = \frac{{ - 2ax}}{{{{\left( {{a^2} + {x^2}} \right)}^2} - {{\left( {ax} \right)}^2}}} + \frac{{2ax}}{{{a^4} + {x^4} + {a^2}{x^2}}} \cr & = \frac{{ - 2ax}}{{{a^4} + {x^4} + 2{a^2}{x^2} - {a^2}{x^2}}} + \frac{{2ax}}{{{a^4} + {x^4} + {a^2}{x^2}}} \cr & = \frac{{ - 2ax}}{{{a^4} + {x^4} + {a^2}{x^2}}} + \frac{{2ax}}{{{a^4} + {x^4} + {a^2}{x^2}}} \cr & = 0 \cr} $$