Question

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 $a + b + c.$

Answer

Answer: 98

Concepts:polynomial rationalizing denominator algebraic manipulation

Difficulty rating: 2400

Solution:

Rewrite the equation as $9x^3 = x^3 + 3x^2 + 3x + 1$ $= (x + 1)^3.$ Taking real cube roots, $\sqrt[3]{9}\,x = x + 1,$ so $x = \frac{1}{\sqrt[3]{9} - 1}.$

Multiply numerator and denominator by $\sqrt[3]{81} + \sqrt[3]{9} + 1;$ the denominator becomes $(\sqrt[3]{9})^3 - 1 = 8,$ so $x = \frac{\sqrt[3]{81} + \sqrt[3]{9} + 1}{8}.$ Thus $a + b + c = 81 + 9 + 8 = 98.$

2013 AIME I Problem 5

Answer: 98

Difficulty rating: 2400

Solution:

Problem 5 in Other Years