Cubic Equation Real Root
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 $.
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- $\frac{a}{b}$
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- $a^n$
- $a^{\circ}$
- $a_n$
- $\sqrt{}$
- $\sqrt[n]{}$
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Solution
In general, the solutions to cubic equations are very messy, so we hope there is a trick to solving this particular equation.
Noticing the $ (3, 3, 1) $ pattern of coefficients, which appears in the expansion \[(x+1)^3 = x^3 + 3x^2 + 3x + 1,\]we rewrite the left-hand side as \[9x^3 - (x^3 + 3x^2 + 3x + 1) = 0\]or \[9x^3 - (x+1)^3 = 0.\]Thus, $ 9x^3 = (x+1)^3 $, and since $ x $ is real, \[x\sqrt[3]{9} = x+1 \implies x =\frac{1}{\sqrt[3]{9}-1}.\]To rationalize the denominator, we write \[x = \frac{1}{\sqrt[3]{9}-1} \cdot \frac{\sqrt[3]{81} + \sqrt[3]{9} + 1}{\sqrt[3]{81} + \sqrt[3]{9} + 1} = \frac{\sqrt[3]{81} + \sqrt[3]{9} + 1}{8}\]by the difference of cubes factorization. The answer is $ 81 + 9 + 8 = \boxed{98} $.
Freeflows