Difference between revisions of "2013 AIME I Problems/Problem 5"
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The real root of the equation <math>8x^3 - 3x^2 - 3x - 1 = 0</math> can be written in the form <math>\frac{\sqrt[3]a + \sqrt[3]b + 1}{c}</math>, where <math>a</math>, <math>b</math>, and <math>c</math> are positive integers. Find <math>a+b+c</math>. | The real root of the equation <math>8x^3 - 3x^2 - 3x - 1 = 0</math> can be written in the form <math>\frac{\sqrt[3]a + \sqrt[3]b + 1}{c}</math>, where <math>a</math>, <math>b</math>, and <math>c</math> are positive integers. Find <math>a+b+c</math>. | ||
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− | + | == Solution 1 == | |
We note that <math>8x^3 - 3x^2 - 3x - 1 = 9x^3 - x^3 - 3x^2 - 3x - 1 = 9x^3 - (x + 1)^3</math>. Therefore, we have that <math>9x^3 = (x+1)^3</math>, so it follows that <math>x\sqrt[3]{9} = x+1</math>. Solving for <math>x</math> yields <math>\frac{1}{\sqrt[3]{9}-1} = \frac{\sqrt[3]{81}+\sqrt[3]{9}+1}{8}</math>, so the answer is <math>\boxed{098}</math>. | We note that <math>8x^3 - 3x^2 - 3x - 1 = 9x^3 - x^3 - 3x^2 - 3x - 1 = 9x^3 - (x + 1)^3</math>. Therefore, we have that <math>9x^3 = (x+1)^3</math>, so it follows that <math>x\sqrt[3]{9} = x+1</math>. Solving for <math>x</math> yields <math>\frac{1}{\sqrt[3]{9}-1} = \frac{\sqrt[3]{81}+\sqrt[3]{9}+1}{8}</math>, so the answer is <math>\boxed{098}</math>. | ||
− | + | == Solution 2 == | |
Let <math>r</math> be the real root of the given [[polynomial]]. Now define the cubic polynomial <math>Q(x)=-x^3-3x^2-3x+8</math>. Note that <math>1/r</math> must be a root of <math>Q</math>. However we can simplify <math>Q</math> as <math>Q(x)=9-(x+1)^3</math>, so we must have that <math>(\frac{1}{r}+1)^3=9</math>. Thus <math>\frac{1}{r}=\sqrt[3]{9}-1</math>, and <math>r=\frac{1}{\sqrt[3]{9}-1}</math>. We can then multiply the numerator and denominator of <math>r</math> by <math>\sqrt[3]{81}+\sqrt[3]{9}+1</math> to rationalize the denominator, and we therefore have <math>r=\frac{\sqrt[3]{81}+\sqrt[3]{9}+1}{8}</math>, and the answer is <math>\boxed{098}</math>. | Let <math>r</math> be the real root of the given [[polynomial]]. Now define the cubic polynomial <math>Q(x)=-x^3-3x^2-3x+8</math>. Note that <math>1/r</math> must be a root of <math>Q</math>. However we can simplify <math>Q</math> as <math>Q(x)=9-(x+1)^3</math>, so we must have that <math>(\frac{1}{r}+1)^3=9</math>. Thus <math>\frac{1}{r}=\sqrt[3]{9}-1</math>, and <math>r=\frac{1}{\sqrt[3]{9}-1}</math>. We can then multiply the numerator and denominator of <math>r</math> by <math>\sqrt[3]{81}+\sqrt[3]{9}+1</math> to rationalize the denominator, and we therefore have <math>r=\frac{\sqrt[3]{81}+\sqrt[3]{9}+1}{8}</math>, and the answer is <math>\boxed{098}</math>. | ||
− | + | == Solution 3 == | |
It is clear that for the algebraic degree of <math>x</math> to be <math>3</math> that there exists some cubefree integer <math>p</math> and positive integers <math>m,n</math> such that <math>a = m^3p</math> and <math>b = n^3p^2</math> (it is possible that <math>b = n^3p</math>, but then the problem wouldn't ask for both an <math>a</math> and <math>b</math>). Let <math>f_1</math> be the [[automorphism]] over <math>\mathbb{Q}[\sqrt[3]{a}][\omega]</math> which sends <math>\sqrt[3]{a} \to \omega \sqrt[3]{a}</math> and <math>f_2</math> which sends <math>\sqrt[3]{a} \to \omega^2 \sqrt[3]{a}</math> (note : <math>\omega</math> is a cubic [[Roots of unity|root of unity]]). | It is clear that for the algebraic degree of <math>x</math> to be <math>3</math> that there exists some cubefree integer <math>p</math> and positive integers <math>m,n</math> such that <math>a = m^3p</math> and <math>b = n^3p^2</math> (it is possible that <math>b = n^3p</math>, but then the problem wouldn't ask for both an <math>a</math> and <math>b</math>). Let <math>f_1</math> be the [[automorphism]] over <math>\mathbb{Q}[\sqrt[3]{a}][\omega]</math> which sends <math>\sqrt[3]{a} \to \omega \sqrt[3]{a}</math> and <math>f_2</math> which sends <math>\sqrt[3]{a} \to \omega^2 \sqrt[3]{a}</math> (note : <math>\omega</math> is a cubic [[Roots of unity|root of unity]]). | ||
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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 | 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> | <cmath>c^3x^3-3c^2x^2-(3c\sqrt[3]{ab}-3c)x-(a+b+1-3\sqrt[3]{ab})=0.</cmath> |
Latest revision as of 19:05, 24 January 2021
Problem
The real root of the equation can be written in the form , where , , and are positive integers. Find .
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
Video Solution
https://www.youtube.com/watch?v=9way8JrtD04&t=555s
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 | ||
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