Reciprocal of Difference of Squares
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Theorem
Reciprocal of Difference of Squares as Sum of Reciprocals
- $\dfrac 1 {x^2 - y^2} = \dfrac 1 {2 x \paren {x + y} } + \dfrac 1 {2 x \paren {x - y} }$
Reciprocal of Difference of Squares as Difference of Reciprocals
- $\dfrac 1 {x^2 - y^2} = \dfrac 1 {2 y \paren {x - y} } - \dfrac 1 {2 y \paren {x + y} }$