# $x = \sqrt[3]{3} + \frac{1}{\sqrt[3]{3}}$, what is $3x^3 - 9x$?

Suppose $x = \sqrt[3]{3} + \frac{1}{\sqrt[3]{3}}$. Then what is $3x^3 - 9x$?

I tried factorizing $3x^3 - 9x = 3x(x^2 - 3)$ then substituting the values which gives me something very lengthy. I eventually got to the answer — $10$ — after working a bit. Are there some good shortcuts?

This is not homework; I'm preparing for an examination.

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$x^3=\left(\sqrt[3]{3} + \frac{1}{\sqrt[3]{3}}\right)^3$

$=\left(\sqrt[3]{3}\right)^3+\left(\frac{1}{\sqrt[3]{3}}\right)^3+3\cdot\sqrt[3]{3}\cdot\frac{1}{\sqrt[3]{3}}\left(\sqrt[3]{3} + \frac{1}{\sqrt[3]{3}}\right)$

$=3+\frac13+3\cdot\sqrt[3]{3}\cdot \frac{1}{\sqrt[3]{3}}\cdot x$ as $\sqrt[3]{3} + \frac{1}{\sqrt[3]{3}}=x$

$=\frac{10}3+3x$

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Ah, so $3x^3 = 10 + 9x$ and I want to get the value of $3x^3 - 9x$, which gives me $10 + 9x - 9x = 10$. Great solution! –  Parth Kohli Apr 13 '13 at 4:49
@pen, this is a way to rationalize $$x = \sqrt[3]a +\frac1{\sqrt[3]a}$$ –  lab bhattacharjee Apr 13 '13 at 4:53

Your method works fine, and is quite simple:

$$\rm x\, =\, a\!+\!a^{-1}\Rightarrow\ x\,(x^2\!-3)\, =\, (a+ a^{-1})(a^2\! -1 + a^{-2})\, =\, a^3\!+ a^{-3}\! = 3 + 3^{-1}$$

Therefore $\rm\ \ 3x(x^2\!-3)\, =\, 3(3+3^{-1})\, =\, 10\ \ \$ QED

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$x^2 - 3 = (a^2 + 2 + a^{-2}) - 3 = a^2 - 1 + a^{-2} \neq a^2 + 1 + a^{-1}$ –  TMM Apr 13 '13 at 23:15
@TMM Yes, that's what was intended. Typos now fixed, thanks. –  Math Gems Apr 13 '13 at 23:27
Another way: Since $x= 3^\frac{1}{3} + 3^\frac{-1}{3}$, $$\Rightarrow x - 3^\frac{-1}{3} = 3^\frac{1}{3}$$ Take cubes on both sides $$x^3 - \frac{1}{3} -3x.3^\frac{-1}{3}(x-3^\frac{-1}{3}) = 3$$
$$x^3 - \frac{1}{3} -3x.3^\frac{-1}{3}3^\frac{1}{3} = 3$$
$$x^3 - \frac{1}{3} -3x = 3$$