# Sum of Fourth Powers with Product of Squares

## 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

$\blacksquare$