Question

plz solve both the questions properly... i would be grateful​

Answer 1

Question -

Find-

$\frac{1}{ {a}^{2} + ax + {x}^{2} } - \frac{1}{ {a}^{2} - ax + {x}^{2} } + \frac{2ax}{ {x}^{4} + {a}^{2} {x}^{2} + {a}^{4} }$

Solution -

$= \frac{1}{ {a}^{2} + ax + {x}^{2} } - \frac{1}{ {a}^{2} - ax + {x}^{2} } + \frac{2ax}{ {x}^{4} + {a}^{2} {x}^{2} + {a}^{4} }$

Taking L. C. M of a²+ax+x² and a²-ax+x² we will get -

$= \frac{ {a}^{2} - ax + {x}^{2} - {a}^{2} - ax - {x}^{2} }{( {a}^{2} + ax + {x}^{2})( {a}^{2} - ax + {x}^{2}) } + \frac{2ax}{ {x}^{4} + {a}^{2} {x}^{2} + {a}^{4} }$

$= \frac{ - 2ax}{ {a}^{2}( {a}^{2} - ax + {x}^{2}) + ax( {a}^{2} - ax + {x}^{2} ) + {x}^{2}( {a}^{2} - ax + {x}^{2} )} + \frac{2ax}{ {x}^{4} + {a}^{2} {x}^{2} + {a}^{4} }$

$= \frac{ - 2ax}{ {a}^{4} - {a}^{3}x + {a}^{2} {x}^{2} + {a}^{3}x - {a}^{2} {x}^{2} + a {x}^{3} + {a}^{2} {x}^{2} - a {x}^{3} + {x}^{4} } + \frac{2ax}{ {x}^{4} + {a}^{2} {x}^{2} + {a}^{4} }$

$= \frac{ - 2ax}{ {x}^{4} + {a}^{2} {x}^{2} + {a}^{4} } + \frac{2ax}{ {x}^{4} + {a}^{2} {x}^{2} + {a}^{4} }$

Taking L. C. M of x⁴+a²x²+a⁴ and x⁴+a²x²+a⁴ we will get -

$= \frac{ - 2ax + 2ax}{( {x}^{4} + {a}^{2} {x}^{2} + {a}^{4} )( {x}^{4} + {a}^{2} {x}^{2} + {a}^{4}) }$

$= \frac{0}{( {x}^{4} + {a}^{2} {x}^{2} + {a}^{4})( {x}^{4} + {a}^{2} {x}^{2} + {a}^{4} )}$

$= 0$

Therefore the answer is 0