Sum of Two Odd Powers/Examples/Sum of Two Cubes/Proof 1
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Theorem
- $x^3 + y^3 = \paren {x + y} \paren {x^2 - x y + y^2}$
Proof
From Difference of Two Powers:
- $\ds a^n - b^n = \paren {a - b} \sum_{j \mathop = 0}^{n - 1} a^{n - j - 1} b^j$
Let $x = a$ and $y = -b$.
Then:
\(\ds x^3 + y^3\) | \(=\) | \(\ds x^3 - \paren {-y^3}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds x^3 - \paren {-y}^3\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \paren {x - \paren {-y} } \sum_{j \mathop = 0}^2 x^{n - j - 1} \paren {-y}^j\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \paren {x + y} \paren {x^2 + x \paren {-y} + \paren {-y}^2}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \paren {x + y} \paren {x^2 - x y + y^2}\) |
$\blacksquare$