What Is the Exact Value of the Real Root in the Equation \(x^3 + 3x - 2 = 0\)?

[anemone]

Here is this week's POTW:

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Find the exact value for the real root of the equation $x^3+3x-2=0$.

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Remember to read the http://www.mathhelpboards.com/showthread.php?772-Problem-of-the-Week-%28POTW%29-Procedure-and-Guidelines to find out how to http://www.mathhelpboards.com/forms.php?do=form&fid=2!

 

[anemone]

Congratulations to the following members for their correct solution::)

1. Opalg
2. MarkFL
3. lfdahl

Solution from Opalg:

Spoiler

Using Vieta's method, let $x = s - s^{-1}$. Then $$0 = x^3 + 3x - 2 = (s - s^{-1})^3 + 3(s - s^{-1}) - 2 = s^3 - s^{-3} - 2.$$ After multiplying by $s^3$, this becomes a quadratic equation $s^6 - 2s^3 - 1 = 0$ in $s^3$, with solutions $s^3 = 1 \pm\sqrt2$. Taking the larger root (the other one would lead to the same result) gives $s = \bigl(\sqrt2 + 1\bigr)^{1/3}$, and $x = s - s^{-1} = \bigl(\sqrt2 + 1\bigr)^{1/3} - \bigl(\sqrt2 + 1\bigr)^{-1/3}.$