Reciprocal of Difference of Squares as Difference of Reciprocals

Theorem

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

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

$=$

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

putting everything over a common denominator

$\ds $

$=$

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

$\ds \leadsto \ \ $

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

$=$

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

dividing both sides by $2 y$

$\blacksquare$