Difference of Two Fourth Powers

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Theorem

$x^4 - y^4 = \paren {x - y} \paren {x + y} \paren {x^2 + y^2}$


Proof 1

\(\ds x^4 - y^4\) \(=\) \(\ds \paren {x^2}^2 - \paren {y^2}^2\)
\(\ds \) \(=\) \(\ds \paren {x^2 - y^2} \paren {x^2 + y^2}\) Difference of Two Squares
\(\ds \) \(=\) \(\ds \paren {x - y} \paren {x + y} \paren {x^2 + y^2}\) Difference of Two Squares

$\blacksquare$


Proof 2

\(\ds x^4 - y^4\) \(=\) \(\ds \paren {x - y} \paren {x^3 + x^2 y + x y^2 + y^3}\) Difference of Two Powers
\(\ds \) \(=\) \(\ds \paren {x - y} \paren {\paren {x^3 + y^3} + x y \paren {x + y} }\) rearranging
\(\ds \) \(=\) \(\ds \paren {x - y} \paren {\paren {x + y} \paren {x^2 - x y + y^2} + x y \paren {x + y} }\) Sum of Two Cubes
\(\ds \) \(=\) \(\ds \paren {x - y} \paren {x + y} \paren {\paren {x^2 - x y + y^2} + x y}\) extracting common factor
\(\ds \) \(=\) \(\ds \paren {x - y} \paren {x + y} \paren {x^2 + y^2}\) simplifying

$\blacksquare$


Sources