# Thread: need help in exponent problem

1. ## need help in exponent problem

if x+1/x = 3^1/2 then x^3 - 1/x^3 = ??

thanks

2. $\displaystyle \displaystyle\frac{x+1}{x}=\sqrt{3}$

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

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

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

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

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

Surely, you can take it from here.

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

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

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

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

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

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

The .... are left for you to think about and fill in.

4. Hello, disharmonie!

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

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

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

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

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

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

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

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

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

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

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