Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
up vote 6 down vote accepted

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

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

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

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

share|cite|improve this answer
    
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

share|cite|improve this answer
    
$x^2 - 3 = (a^2 + 2 + a^{-2}) - 3 = a^2 - 1 + a^{-2} \neq a^2 + 1 + a^{-1}$ – 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 $$ x^3 - \frac{1}{3} -3x.3^\frac{-1}{3}(x-3^\frac{-1}{3}) = 3$$

$$ x^3 - \frac{1}{3} -3x.3^\frac{-1}{3}3^\frac{1}{3} = 3$$

$$x^3 - \frac{1}{3} -3x = 3$$

Multiply by 3 to get your answer :)

share|cite|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.