Difference of Two Sixth Powers
Theorem
- $x^6 - y^6 = \paren {x - y} \paren {x + y} \paren {x^2 + x y + y^2} \paren {x^2 - x y + y^2}$
Proof
| \(\ds x^6 - y^6\) | \(=\) | \(\ds \paren {x^3}^2 - \paren {y^3}^2\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \paren {x^3 - y^3} \paren {x^3 + y^3}\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \paren {x - y} \paren {x^2 + x y + y^2} \paren {x^3 + y^3}\) | Difference of Two Cubes | |||||||||||
| \(\ds \) | \(=\) | \(\ds \paren {x - y} \paren {x^2 + x y + y^2} \paren {x + y} \paren {x^2 - x y + y^2}\) | Sum of Two Cubes |
$\blacksquare$
Sources
- 1968: Murray R. Spiegel: Mathematical Handbook of Formulas and Tables ... (previous) ... (next): $\S 2$: Special Products and Factors: $2.17$
- 2009: Murray R. Spiegel, Seymour Lipschutz and John Liu: Mathematical Handbook of Formulas and Tables (3rd ed.) ... (previous) ... (next): $\S 2$: Special Products and Factors: $2.17.$