Difference of Two Fourth Powers
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Theorem
- $x^4 - y^4 = \paren {x - y} \paren {x + y} \paren {x^2 + y^2}$
Proof 1
\(\ds x^4 - y^4\) | \(=\) | \(\ds \paren {x^2}^2 - \paren {y^2}^2\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \paren {x^2 - y^2} \paren {x^2 + y^2}\) | Difference of Two Squares | |||||||||||
\(\ds \) | \(=\) | \(\ds \paren {x - y} \paren {x + y} \paren {x^2 + y^2}\) | Difference of Two Squares |
$\blacksquare$
Proof 2
\(\ds x^4 - y^4\) | \(=\) | \(\ds \paren {x - y} \paren {x^3 + x^2 y + x y^2 + y^3}\) | Difference of Two Powers | |||||||||||
\(\ds \) | \(=\) | \(\ds \paren {x - y} \paren {\paren {x^3 + y^3} + x y \paren {x + y} }\) | rearranging | |||||||||||
\(\ds \) | \(=\) | \(\ds \paren {x - y} \paren {\paren {x + y} \paren {x^2 - x y + y^2} + x y \paren {x + y} }\) | Sum of Two Cubes | |||||||||||
\(\ds \) | \(=\) | \(\ds \paren {x - y} \paren {x + y} \paren {\paren {x^2 - x y + y^2} + x y}\) | extracting common factor | |||||||||||
\(\ds \) | \(=\) | \(\ds \paren {x - y} \paren {x + y} \paren {x^2 + y^2}\) | simplifying |
$\blacksquare$
Sources
- 1968: Murray R. Spiegel: Mathematical Handbook of Formulas and Tables ... (previous) ... (next): $\S 2$: Special Products and Factors: $2.14$