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: $2.17$