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
- 1968: Murray R. Spiegel: Mathematical Handbook of Formulas and Tables ... (previous) ... (next): $\S 2$: Special Products and Factors: $2.18$