SOLUTION TO PROBLEM 4

It follows from the Binomial Theorem of Newton, that

$$\begin{eqnarray*}(a+b)^2& =&a^2+2ab+b^2,\\ (a+b)^3& =&a^3+3a^2b+3ab^2+b^3,\\ (a+b)^5& =&a^5+5a^4b+10a^3b^2+10a^2b^3+5ab^4+b^5. \end{eqnarray*}$$

Since c=-a-b, we find that

$$\begin{eqnarray*}\frac{a^2+b^2+c^2}{2} &=& a^2+ab+b^2, \\ \frac{a^3+b^3+c^3}{3}... ...ab^2),\\ \frac{a^5+b^5+c^5}{5}&=& -(a^4b+2a^3b^2+2a^2b^3+ab^4). \end{eqnarray*}$$

By verifying

$$(a^2+ab+b^2)\cdot (a^2b+ab^2)= a^4b+2a^3b^2+2a^2b^3+ab^4$$

the identity is proven.

Questions and/or comments should be directed to Judy Downey or Griff Elder

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Last modified:   Fri Feb 2 19:32:17 CST 2001