Find all real numbers $x$, $y$, $z$, $t$, $u$ that
$x+y+z+u+t=0$
$x^3+y^3+z^3+u^3+t^3=0$
$x^5+y^5+z^5+u^5+t^5=-10$
I'm learning about Chebyshev polynominals but in this case, I still haven't got any idea :(( please help me
Solving $x+y+z+u+t=0$, $x^3+y^3+z^3+u^3+t^3=0$, $x^5+y^5+z^5+u^5+t^5=-10$
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$\begingroup$
polynomials
systems-of-equations
roots
symmetric-polynomials
chebyshev-polynomials
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0what kind of numbers are this? – 2017-01-20
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0they are real of cause – 2017-01-20
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1I am not sure whether it helps, but there is a question about very similar system, but with the restriction that the 5 numbers are from the interval $[-2,2]$: [A system of equations with 5 variables: $a+b+c+d+e=0$, $a^3+b^3+c^3+d^3+e^3=0$, $a^5+b^5+c^5+d^5+e^5=10$](http://math.stackexchange.com/q/162854), Found [using Approach0](https://approach0.xyz/search/?q=%24x%2By%2Bz%2Bu%2Bt%3D0%24%2C%20%24x%5E3%2By%5E3%2Bz%5E3%2Bu%5E3%2Bt%5E3%3D0%24%2C%20%24x%5E5%2By%5E5%2Bz%5E5%2Bu%5E5%2Bt%5E5%3D-10%24&p=1). – 2017-01-21
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0thanks, it's helped me so much :)) – 2017-01-21
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0however, I still have one more question, that when we know that the sum of cos5a = 5, how to do next – 2017-01-21
1 Answers
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There are many real solutions. To obtain examples, we choose, say, $x=-2$ and $y=3$. Taking resultants we obtain the equation $$ 3u^3+3u^2-19u-1=0, $$ and $z$ as a root of a quadratic equation in $u$. Now there are three real solutions, e.g., $u=2.09659797695$. Then $$ (x,y,z,u,t)=(-2,3,- 3.04437453374,2.09659797695,- 0.0522234432082) $$ is a solution.
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0yes that right, but what we need to do here is figure out the form of each number – 2017-01-20
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0of course with 5 unknowns and just 3 equations, we can't do more – 2017-01-20
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0You could edit the question to say more on what we need to do here. – 2017-01-20