Difference between revisions of "2013 AIME I Problems/Problem 5"
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Now, note that <math>\sqrt[3]{a} + \sqrt[3]{b} + 1</math> is a root of <math>x^3 - 3x^2 - 24x - 64 = 0</math>. Thus <math>(x-1)^3 = 27x + 63</math> so <math>(\sqrt[3]{a} + \sqrt[3]{b})^3 = 27(\sqrt[3]{a} + \sqrt[3]{b}) + 90</math>. Checking the non-cubicroot dimension part, we get <math>a + b = 90</math> so it follows that <math>a + b + c = \boxed{098}</math>. | Now, note that <math>\sqrt[3]{a} + \sqrt[3]{b} + 1</math> is a root of <math>x^3 - 3x^2 - 24x - 64 = 0</math>. Thus <math>(x-1)^3 = 27x + 63</math> so <math>(\sqrt[3]{a} + \sqrt[3]{b})^3 = 27(\sqrt[3]{a} + \sqrt[3]{b}) + 90</math>. Checking the non-cubicroot dimension part, we get <math>a + b = 90</math> so it follows that <math>a + b + c = \boxed{098}</math>. | ||
+ | === Solution 4 === | ||
+ | We have <math>cx-1=\sqrt[3]{a}+\sqrt[3]{b}.</math> Therefore <math>(cx-1)^3=(\sqrt[3]{a}+\sqrt[3]{b})^3=a+b+3\sqrt[3]{ab}(\sqrt[3]{a}+\sqrt[3]{b})=a+b+3\sqrt[3]{ab}(cx-1).</math> We have | ||
+ | <cmath>c^3x^3-3c^2x^2-(3c\sqrt[3]{ab}-3c)x-(a+b+1-3\sqrt[3]{ab})=0.</cmath> | ||
+ | We will find <math>a,b,c</math> so that the equation is equivalent to the original one. Let <math>\dfrac{3c^2}{c^3}=\dfrac{3}{8}, \dfrac{3c\sqrt[3]{ab}-3c}{c^3}=\dfrac{3}{8}, \dfrac{a+b+1-3\sqrt[3]{ab}}{c^3}=\dfrac{1}{8}.</math> Easily, <math>c=8, \sqrt[3]{ab}=9,</math> and <math>a+b=90.</math> So <math>a + b + c = 90+8=\boxed{098}</math>. | ||
+ | -JZ | ||
== See Also == | == See Also == |
Revision as of 10:52, 15 February 2020
Problem
The real root of the equation can be written in the form
, where
,
, and
are positive integers. Find
.
Contents
Solutions
Solution 1
We note that . Therefore, we have that
, so it follows that
. Solving for
yields
, so the answer is
.
Solution 2
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
.
Solution 3
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
.
Solution 4
We have Therefore
We have
We will find
so that the equation is equivalent to the original one. Let
Easily,
and
So
.
-JZ
See Also
2013 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 4 |
Followed by Problem 6 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.