2013 AIME I Problems/Problem 5

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 $a+b+c$ .

Solutions

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 $\boxed{098}$ .

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

2013 AIME I (Problems • Answer Key • Resources)
Preceded by Problem 4	Followed by Problem 6
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