Summation of Powers over Product of Differences/Example

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Theorem

\(\ds \frac 1 {\paren {a - b} \paren {a - c} } + \frac 1 {\paren {b - a} \paren {b - c} } + \frac 1 {\paren {c - a} \paren {c - b} }\) \(=\) \(\ds 0\)
\(\ds \frac a {\paren {a - b} \paren {a - c} } + \frac b {\paren {b - a} \paren {b - c} } + \frac c {\paren {c - a} \paren {c - b} }\) \(=\) \(\ds 0\)
\(\ds \frac {a^2} {\paren {a - b} \paren {a - c} } + \frac {b^2} {\paren {b - a} \paren {b - c} } + \frac {c^2} {\paren {c - a} \paren {c - b} }\) \(=\) \(\ds 1\)
\(\ds \frac {a^3} {\paren {a - b} \paren {a - c} } + \frac {b^3} {\paren {b - a} \paren {b - c} } + \frac {c^3} {\paren {c - a} \paren {c - b} }\) \(=\) \(\ds a + b + c\)


Proof

Instance of Summation of Powers over Product of Differences:

$\ds \sum_{j \mathop = 1}^n \paren {\dfrac { {x_j}^r} {\ds \prod_{\substack {1 \mathop \le k \mathop \le n \\ k \mathop \ne j} } \paren {x_j - x_k} } } = \begin {cases} 0 & : 0 \le r < n - 1 \\ 1 & : r = n - 1 \\ \ds \sum_{j \mathop = 1}^n x_j & : r = n \end {cases}$

where:

$n = 3$
$x_1 = a$
$x_2 = b$
$x_3 = c$

$\blacksquare$


Sources

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