1. ## Any guideline??

The problem
$\displaystyle (1 - x^-2) : (1 - x^-1)$

My aproach:

= $\displaystyle (\frac{1}{1} - \frac{1}{x^2}) : (\frac{1}{1} - \frac{1}{x})$

= $\displaystyle (\frac{x^2}{x^2} - \frac{1}{x^2}) : (\frac{x}{x} - \frac{1}{x})$

= $\displaystyle \frac{x^2 - 1}{x^2} : \frac{x - 1}{x}$

= $\displaystyle \frac{x^2 - 1}{x^2} * \frac{x}{x - 1}$

= $\displaystyle \frac{x^3 - x}{x^3 - x^2}$

But according to the book the answer should be = $\displaystyle \frac{x + 1}{x}$

Mmm...-2 and -1 on the original problem should be exponents.

2. Originally Posted by Alienis Back
The problem
$\displaystyle (1 - x^-2) : (1 - x^-1)$

My aproach:

= $\displaystyle (\frac{1}{1} - \frac{1}{x^2}) : (\frac{1}{1} - \frac{1}{x})$

= $\displaystyle (\frac{x^2}{x^2} - \frac{1}{x^2}) : (\frac{x}{x} - \frac{1}{x})$

= $\displaystyle \frac{x^2 - 1}{x^2} : \frac{x - 1}{x}$

= $\displaystyle \frac{x^2 - 1}{x^2} * \frac{x}{x - 1}$

= $\displaystyle \frac{x^3 - x}{x^3 - x^2}$

But according to the book the answer should be = $\displaystyle \frac{x + 1}{x}$

Mmm...-2 and -1 on the original problem should be exponents.
You haven't gone far enough.

$\displaystyle \frac{x^3 - x}{x^3 - x^2}$

$\displaystyle = \frac{x(x^2 - 1)}{x^2(x - 1)}$

$\displaystyle = \frac{x^2 - 1}{x(x - 1)}$

$\displaystyle = \frac{(x + 1)(x - 1)}{x(x - 1)}$

$\displaystyle = \frac{x + 1}{x}$.

3. What did you do here?

Originally Posted by Prove It
You haven't gone far enough.

$\displaystyle = \frac{x(x^2 - 1)}{x^2(x - 1)}$

$\displaystyle = \frac{x^2 - 1}{x(x - 1)}$.

I mean $\displaystyle x:x^2 = x^1-2 = x^-1$...mmm...

4. Originally Posted by Alienis Back
What did you do here?

I mean $\displaystyle x:x^2 = x^1-2 = x^-1$...mmm...
[tex]Remember that $\displaystyle x^2 = x\cdot x$

So $\displaystyle \frac{x}{x^2} = \frac{x}{x\cdot x} = \frac{1}{x}$.

Therefore

$\displaystyle \frac{x(x^2 - 1)}{x^2(x - 1)} = \frac{x(x^2 - 1)}{x\cdot x(x - 1)} = \frac{1(x^2 - 1)}{x(x - 1)}$