Difference between revisions of "2007 UNCO Math Contest II Problems/Problem 4"

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== Solution ==
 
== Solution ==
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<math>\fbox{+1}</math>
  
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Since <math>x^3=1</math> ,<math>x^{2006}=x^2</math> , <math>x^{2007}=1</math>, <math>x+\frac{1}{x}=-1</math>
  
 
== See Also ==
 
== See Also ==

Latest revision as of 05:11, 12 January 2019

Problem

If $x$ is a primitive cube root of one (this means that $x^3 =1$ but $x \ne 1$) compute the value of \[x^{2006}+\frac{1}{x^{2006}}+x^{2007}+\frac{1}{x^{2007}}.\]

Solution

$\fbox{+1}$

Since $x^3=1$ ,$x^{2006}=x^2$ , $x^{2007}=1$, $x+\frac{1}{x}=-1$

See Also

2007 UNCO Math Contest II (ProblemsAnswer KeyResources)
Preceded by
Problem 3
Followed by
Problem 5
1 2 3 4 5 6 7 8 9 10
All UNCO Math Contest Problems and Solutions
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