# need help in exponent problem

• January 30th 2011, 11:09 AM
disharmonie
need help in exponent problem
if x+1/x = 3^1/2 then x^3 - 1/x^3 = ??

thanks :)
• January 30th 2011, 11:40 AM
rtblue
$\displaystyle\frac{x+1}{x}=\sqrt{3}$

note that 3^(1/2)=sqrt(3).

$\displaystyle x+1=x\sqrt{3}$

$\displaystyle 1=x\sqrt{3}-x$

$\displaystyle 1=x(\sqrt{3}-1)$

$\displaystyle x=\frac{1}{\sqrt{3}-1}$

Surely, you can take it from here.
• January 30th 2011, 11:43 AM
mr fantastic
Quote:

Originally Posted by disharmonie
if x+1/x = 3^1/2 then x^3 - 1/x^3 = ??

thanks :)

Let $p = x + \frac{1}{x} = \sqrt{3}$.

Then $p^2 = x^2 + \frac{1}{x^2} + 2 \Rightarrow x^2 + \frac{1}{x} = .....$.

Let $q = x - \frac{1}{x}$.

Then $q^2 = ..... = ..... \Rightarrow q = .....$

Now note that $x^3 - \frac{1}{x^3} = \left(x^2 + \frac{1}{x^2}\right)q$

The .... are left for you to think about and fill in.
• January 30th 2011, 06:59 PM
Soroban
Hello, disharmonie!

I got a rather silly answer . . . or is it silly?

Quote:

$\displaystyle \text{If }x+\frac{1}{x} \:=\:\sqrt{3},\,\text{ then }\,x^3 - \frac{1}{x^3} \:=\: ?$

$\displaystyle\text{We have: }\;x + \frac{1}{x} \:=\:\sqrt{3}$

$\displaystyle \text{Square both sides: }\;\left(x + \frac{1}{x}\right)^2 \:=\:\left(\sqrt{3}\right)^2 \quad\Rightarrow\quad x^2 + 2 + \frac{1}{x^2} \;=\;3
$

. . $\displaystyle\text{Hence: }\;\boxed{x^2 + \frac{1}{x^2} \;=\;1}$

$\displaystyle \text{Subtract 2 from both sides: }\;x^2 - 2 + \frac{1}{x^2} \;=\;-1 \quad\Rightarrow\quad \left(x - \frac{1}{x}\right)^2 \:=\:-1$

. . $\displaystyle \text{Hence: }\;\boxed{x - \frac{1}{x} \;=\;\pm i}$

$\displaystyle\text{Factor: }\;x^3 - \frac{1}{x^3} \;=\;\left(x - \frac{1}{x}\right)\left(x^2 + 1 + \frac{1}{x^2}\right)$

. . . . . . . . . . . . . $\displaystyle = \;\underbrace{\left(x - \frac{1}{x}\right)}_{\text{This is }\pm i} \left(\underbrace{x^2 + \frac{1}{x^2}}_{\text{This is 1}} + 1\right)$

. . . . . . . . . . . . . $=\;\;(\pm i)(2)$

. . . . . . . . . . . . . $=\;\;\pm 2i$