Reciprocal of Difference of Squares as Sum of Reciprocals

Theorem

$\dfrac 1 {x^2 - y^2} = \dfrac 1 {2 x \paren {x + y} } + \dfrac 1 {2 x \paren {x - y} }$

$\ds \dfrac 1 {x + y} + \dfrac 1 {x - y}$

$=$

$\ds \dfrac {\paren {x - y} + \paren {x + y} } {\paren {x - y} \paren {x + y} }$

putting everything over a common denominator

$\ds $

$=$

$\ds \dfrac {2 x} {x^2 - y^2}$

$\ds \leadsto \ \ $

$\ds \dfrac 1 {2 x \paren {x + y} } + \dfrac 1 {2 x \paren {x - y} }$

$=$

$\ds \dfrac 1 {x^2 - y^2}$

dividing both sides by $2 x$

$\blacksquare$