Difference between revisions of "2007 UNCO Math Contest II Problems/Problem 4"
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== Solution == | == Solution == | ||
| − | {{ | + | <math>\fbox{+1}</math> |
| + | |||
| + | 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 04:11, 12 January 2019
Problem
If 
is a primitive cube root of one (this means that 
but 
) compute the value of
![\[x^{2006}+\frac{1}{x^{2006}}+x^{2007}+\frac{1}{x^{2007}}.\]](http://latex.artofproblemsolving.com/4/d/3/4d3fe43557d68d9c46f840e5241e58392fcc1dd9.png)
Solution
Since 
,
, 
, 
See Also
| 2007 UNCO Math Contest II (Problems • Answer Key • Resources) | ||
| 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 | ||
