Sum of Fourth Powers with Product of Squares

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Theorem

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


Proof

\(\ds x^6 - y^6\) \(=\) \(\ds \paren {x - y} \paren {x + y} \paren {x^2 + 2 x y + 2 y^2} \paren {x^2 - 2 x y + 2 y^2}\) Difference of Two Sixth Powers
\(\ds \) \(=\) \(\ds \paren {x^2 - y^2} \paren {x^2 + 2 x y + 2 y^2} \paren {x^2 - 2 x y + 2 y^2}\) Difference of Two Squares

Then:

\(\ds x^6 - y^6\) \(=\) \(\ds \paren {x^2}^3 - \paren {y^2}^3\)
\(\ds \) \(=\) \(\ds \paren {x^2 - y^2} \paren {\paren {x^2}^2 + \paren {x^2} \paren {y^2} + \paren {y^2}^2}\) Difference of Two Cubes
\(\ds \) \(=\) \(\ds \paren {x^2 - y^2} \paren {x^4 + x^2 y^2 + y^4}\) simplifying

Thus:

\(\ds \paren {x^2 - y^2} \paren {x^4 + x^2 y^2 + y^4}\) \(=\) \(\ds \paren {x^2 - y^2} \paren {x^2 + 2 x y + 2 y^2} \paren {x^2 - 2 x y + 2 y^2}\) as both equal $x^6 - y^6$
\(\ds \leadsto \ \ \) \(\ds \paren {x^4 + x^2 y^2 + y^4}\) \(=\) \(\ds \paren {x^2 + 2 x y + 2 y^2} \paren {x^2 - 2 x y + 2 y^2}\) cancelling $\paren {x^2 - y^2}$ from both sides

$\blacksquare$


Sources