Difference between revisions of "2013 AIME I Problems/Problem 5"

Latest revision as of 20:11, 27 August 2023

Problem

The real root of the equation $8x^3 - 3x^2 - 3x - 1 = 0$ can be written in the form $\frac{\sqrt[3]a + \sqrt[3]b + 1}{c}$ , where $a$ , $b$ , and $c$ are positive integers. Find $a+b+c$ .

Solution 1

We note that $8x^3 - 3x^2 - 3x - 1 = 9x^3 - x^3 - 3x^2 - 3x - 1 = 9x^3 - (x + 1)^3$ . Therefore, we have that $9x^3 = (x+1)^3$ , so it follows that $x\sqrt[3]{9} = x+1$ . Solving for $x$ yields $\frac{1}{\sqrt[3]{9}-1} = \frac{\sqrt[3]{81}+\sqrt[3]{9}+1}{8}$ , so the answer is $\boxed{98}$ .

Solution 2

Let $r$ be the real root of the given polynomial. Now define the cubic polynomial $Q(x)=-x^3-3x^2-3x+8$ . Note that $1/r$ must be a root of $Q$ . However we can simplify $Q$ as $Q(x)=9-(x+1)^3$ , so we must have that $(\frac{1}{r}+1)^3=9$ . Thus $\frac{1}{r}=\sqrt[3]{9}-1$ , and $r=\frac{1}{\sqrt[3]{9}-1}$ . We can then multiply the numerator and denominator of $r$ by $\sqrt[3]{81}+\sqrt[3]{9}+1$ to rationalize the denominator, and we therefore have $r=\frac{\sqrt[3]{81}+\sqrt[3]{9}+1}{8}$ , and the answer is $\boxed{98}$ .

Solution 3

It is clear that for the algebraic degree of $x$ to be $3$ that there exists some cubefree integer $p$ and positive integers $m,n$ such that $a = m^3p$ and $b = n^3p^2$ (it is possible that $b = n^3p$ , but then the problem wouldn't ask for both an $a$ and $b$ ). Let $f_1$ be the automorphism over $\mathbb{Q}[\sqrt[3]{a}][\omega]$ which sends $\sqrt[3]{a} \to \omega \sqrt[3]{a}$ and $f_2$ which sends $\sqrt[3]{a} \to \omega^2 \sqrt[3]{a}$ (note : $\omega$ is a cubic root of unity).

Letting $r$ be the root, we clearly we have $r + f_1(r) + f_2(r) = \frac{3}{8}$ by Vieta's formulas. Thus it follows $c=8$ . Now, note that $\sqrt[3]{a} + \sqrt[3]{b} + 1$ is a root of $x^3 - 3x^2 - 24x - 64 = 0$ . Thus $(x-1)^3 = 27x + 63$ so $(\sqrt[3]{a} + \sqrt[3]{b})^3 = 27(\sqrt[3]{a} + \sqrt[3]{b}) + 90$ . Checking the non-cubicroot dimension part, we get $a + b = 90$ so it follows that $a + b + c = \boxed{98}$ .

Solution 4

We have $cx-1=\sqrt[3]{a}+\sqrt[3]{b}.$ Therefore $(cx-1)^3=(\sqrt[3]{a}+\sqrt[3]{b})^3=a+b+3\sqrt[3]{ab}(\sqrt[3]{a}+\sqrt[3]{b})=a+b+3\sqrt[3]{ab}(cx-1).$ We have $c^3x^3-3c^2x^2-(3c\sqrt[3]{ab}-3c)x-(a+b+1-3\sqrt[3]{ab})=0.$ We will find $a,b,c$ so that the equation is equivalent to the original one. Let $\dfrac{3c^2}{c^3}=\dfrac{3}{8}, \dfrac{3c\sqrt[3]{ab}-3c}{c^3}=\dfrac{3}{8}, \dfrac{a+b+1-3\sqrt[3]{ab}}{c^3}=\dfrac{1}{8}.$ Easily, $c=8, \sqrt[3]{ab}=9,$ and $a+b=90.$ So $a + b + c = 90+8=\boxed{98}$ .

Video Solution

https://www.youtube.com/watch?v=9way8JrtD04&t=240s

@@ Line 1: / Line 1: @@
-==Problem 5==
+__TOC__
+== Problem ==
+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>.
+== 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{98}</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{98}</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]]).
+Letting <math>r</math> be the root, we clearly we have <math>r + f_1(r) + f_2(r) = \frac{3}{8}</math> by Vieta's formulas. Thus it follows <math>c=8</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{98}</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{98}</math>.
+==Video Solution==
+https://www.youtube.com/watch?v=9way8JrtD04&t=240s
+== See Also ==
+{{AIME box|year=2013|n=I|num-b=4|num-a=6}}
+{{MAA Notice}}
+[[Category:Intermediate Algebra Problems]]

2013 AIME I (Problems • Answer Key • Resources)
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