Showing $\frac{x^2+y^2+z^2}{2}\cdot\frac{x^5+y^5+z^5}{5}=\frac{x^7+y^7+z^7}{7}$ when $x+y+z=0$

Asked Aug 06 '22 at 17:37

Active Aug 06 '22 at 17:41

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If $x+y+z=0$ prove that $$\left(\frac{x^2+y^2+z^2}{2}\right)\cdot \left(\frac{x^5+y^5+z^5}{5}\right)=\frac{x^7+y^7+z^7}{7}$$

My work: After doing some calculation, I came on the conclusion that if the above statement is true then this statement $$\frac{x^2(y^5+z^5)+y^2(x^5+z^5)+z^2(y^5+x^5)}{x^7+y^7+z^7}=\frac37$$ must also be true.

Aftet this I'm stuck. Any help is greatly appreciated.

edited Aug 06 '22 at 17:41

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asked Aug 06 '22 at 17:37

@DietrichBurde you really nailed it bro but can you pls give a full expansion – Aug 06 '22 at 17:46
I'm having some problems in determining the denominator – Aug 06 '22 at 17:47
See quora for a solution. – Dietrich Burde Aug 06 '22 at 17:48
Also see https://math.stackexchange.com/q/851985/977780 – Sourav Ghosh Aug 06 '22 at 17:51
@NobodyNobody Please search on MSE before posting. You could find similar questions and even you could find the same question that you are going to ask. – Sourav Ghosh Aug 06 '22 at 17:57

Showing $\frac{x^2+y^2+z^2}{2}\cdot\frac{x^5+y^5+z^5}{5}=\frac{x^7+y^7+z^7}{7}$ when $x+y+z=0$

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