Let $x,y,z \in \mathbb{R}$ such that $xy+yz+zx=0$ and $xyz \neq 0.$ Find the value of $\frac{x^5+y^5+z^5-1}{xyz}.$
=$...\frac{(x+y+z)^5+5xyz(x+y+z)^2−1}{xyz}$
so for x+y+z=1 we have 5. Is it possible to find another value?
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piteer
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1that equal you wrote is comparing which quantities exactly? – Davide Trono Jul 17 '21 at 13:52
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This from the problem. – piteer Jul 17 '21 at 14:07
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If $x+y+z=1$ is not a requirement, then there are tons of other possibilities. For example, the expression will be $-\frac{31}{16}$ when $x=1,y=1,z=-\frac 1 2$. – Pythagoras Jul 17 '21 at 16:39
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"Is it possible to find another value?" Yes. Choose arbitrary values for $x$ and $y$. Solve for $z$. Substitute these into your expression and will you get another value. – M. Wind Jul 17 '21 at 17:18
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Ok, thanks I see that. – piteer Jul 17 '21 at 17:19