2013 AIME I Problems/Problem 5

Revision as of 13:10, 12 August 2013 by Flamefoxx99 (talk | contribs) (Solution 3)


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 have that $9x^3 = (x+1)^3$, so it follows that $\sqrt[3]{9}x = x+1$. Solving for $x$ yields $\frac{1}{\sqrt[3]{9}-1} = \frac{\sqrt[3]{81}+\sqrt[3]{9}+1}{8}$, so the answer is $\boxed{098}$.

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 $\boxed{098}$.

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. 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 $a + b + c = \boxed{098}$.

Solution 4

We proceed by using the cubic formula.

Let $a=8$, $b=-3$, $c=-3$, and $d=-1$. Then let $m=\left(\dfrac{-b^3}{27a^3}+\dfrac{bc}{6a^2}-\dfrac{d}{2a}\right)$ and $n=\left(\dfrac{c}{3a}-\dfrac{b^2}{9a^2}\right)$. Then the real root of $ax^3+bx^2+cx+d$ is \[\sqrt[3]{m+\sqrt{m^2+n^3}}+\sqrt[3]{m-\sqrt{m^2+n^3}}-\dfrac{b}{3a}\] Now note that \[m=\dfrac{27}{27\cdot 512}+\dfrac{3}{128}+\dfrac{1}{16}=\dfrac{1}{512}+\dfrac{12}{512}+\dfrac{32}{512}=\dfrac{45}{512}\] and \[n=\dfrac{-3}{24}-\dfrac{9}{576}=\dfrac{-9}{64}\] Thus \[r=\sqrt[3]{\dfrac{45}{512}+\sqrt{\dfrac{45^2}{512^2}-\dfrac{9^3}{64^3}}}+\sqrt[3]{\dfrac{45}{512}-\sqrt{\dfrac{45^2}{512^2}-\dfrac{9^3}{64^3}}}+\dfrac{3}{24}\] \[=\sqrt[3]{\dfrac{45}{512}+\sqrt{\dfrac{45^2 - 729}{2^{18}}}}+\sqrt[3]{\dfrac{45}{512}-\sqrt{\dfrac{45^2 - 729}{2^{18}}}}+\dfrac{1}{8}\] \[=\sqrt[3]{\dfrac{45}{512}+\dfrac{36}{512}}+\sqrt[3]{\dfrac{45}{512}-\dfrac{36}{512}}+\dfrac{1}{8}\] \[=\dfrac{\sqrt[3]{81}}{8}+\dfrac{\sqrt[3]{9}}{8}+\dfrac{1}{8}=\dfrac{\sqrt[3]{81}+\sqrt[3]{9}+1}{8}\] and hence the answer is $81+9+8=\boxed{098}$.

See Also

2013 AIME I (ProblemsAnswer KeyResources)
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

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