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Question Number 97675 by Farruxjano last updated on 09/Jun/20

	Answered by mr W last updated on 09/Jun/20

	$p_{n} = x^{n} + y^{n} + z^{n}$ $e_{1} = x + y + z$ $e_{2} = x y + y z + z x$ $e_{3} = x y z$ $e_{⩾ 4} = 0$ $p_{1} = e_{1} = 0$ $p_{2} = e_{1} p_{1} - 2 e_{2} = - 2 e_{2}$ $p_{3} = e_{1} p_{2} - e_{2} p_{1} + 3 e_{3} = 3 e_{3}$ $p_{4} = e_{1} p_{3} - e_{2} p_{2} + e_{3} p_{1} = - e_{2} p_{2} = 2 e_{2}^{2}$ $p_{5} = e_{1} p_{4} - e_{2} p_{3} + e_{3} p_{2} = - e_{2} p_{3} + e_{3} p_{2} = - 5 e_{2} e_{3}$ $p_{7} = e_{1} p_{6} - e_{2} p_{5} + e_{3} p_{4} = - e_{2} p_{5} + e_{3} p_{4} = 7 e_{2}^{2} e_{3}$ $\frac{p_{7}}{7} = e_{2}^{2} e_{3}$ $\frac{p_{5}}{5} = - e_{2} e_{3}$ $\frac{p_{2}}{2} = - e_{2}$ $\frac{p_{5}}{5} \times \frac{p_{2}}{2} = (- e_{2} e_{3}) (- e_{2}) = e_{2}^{2} e_{3} = \frac{p_{7}}{7}$ $\Rightarrow \frac{x^{7} + y^{7} + z^{7}}{7} = \frac{x^{5} + y^{5} + z^{5}}{5} \times \frac{x^{2} + y^{2} + z^{2}}{2}$
	Commented by Farruxjano last updated on 09/Jun/20

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