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Help for II-A Algebra 9.18

 
 
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Re: Help for II-A Algebra 9.18
by David Reynoso - Monday, August 12, 2019, 1:17 PM
 

Since all five roots of $x^5 + 2x -1 = 0$ satisfy the equation, for each of them we have that $x_{k}^5 = -2x_{k} + 1$, ($k=1,2,3,4,5$). Using this, what is $x_{1}^5 + x_{2}^5 + x_{3}^5 + x_{4}^5 + x_{5}^5$ equal to?