2013 AIME I Problems/Problem 5
The real root of the equation can be written in the form , where , , and are positive integers. Find .
We note that . Therefore, we have that , so it follows that . Solving for yields , so the answer is .
Let be the real root of the given polynomial. Now define the cubic polynomial . Note that must be a root of . However we can simplify as , so we must have that . Thus , and . We can then multiply the numerator and denominator of by to rationalize the denominator, and we therefore have , and the answer is .
It is clear that for the algebraic degree of to be that there exists some cubefree integer and positive integers such that and (it is possible that , but then the problem wouldn't ask for both an and ). Let be the automorphism over which sends and which sends (note : is a cubic root of unity).
Letting be the root, we clearly we have by Vieta's formulas. Thus it follows . Now, note that is a root of . Thus so . Checking the non-cubicroot dimension part, we get so it follows that .
We have Therefore We have We will find so that the equation is equivalent to the original one. Let Easily, and So .
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