Algebraically simplify: -x^(-1) +1-(x-1)(x^-2) to ((1/x)-1)^2
Remember that multiplying by negative powers is the same as dividing by that positive power.
So you would have the equation of:
$\displaystyle \dfrac{-1}{x}+1-\dfrac{x-1}{x^2}$
Make each of them have a common denominator by multiplying by corresponding values of x:
$\displaystyle = \dfrac{-x}{x^2}+\dfrac{x^2}{x^2}-\dfrac{x-1}{x^2}$
$\displaystyle =\dfrac{-x+x^2-x+1}{x^2}$
$\displaystyle =\dfrac{x^2-2x+1}{x^2}$
$\displaystyle =\dfrac{(1-x)^2}{x^2}$
Note, x^2-2x+1 can be factorised to (x-1)^2 or (1-x)^2. I used (1-x)^2 because it helps in getting the final answer you gave.
$\displaystyle =\left(\dfrac{1-x}{x}\right)^2$
$\displaystyle =(\frac{1}{x}-\frac{x}{x})^2$
$\displaystyle =(\frac{1}{x}-1)^2$
Hello, yess!
$\displaystyle \text{Simplify }\,-x^{-1} +1-(x-1)x^{-2}\:\text{ to }\:\left(\dfrac{1}{x}-1\right)^2$
We have: .$\displaystyle -\dfrac{1}{x} + 1 - \left(\dfrac{x-1}{x^2}\right) \;\;=\;\;-\dfrac{1}{x} + 1 - \left(\dfrac{x}{x^2} - \dfrac{1}{x^2}\right) $
. . . . . . $\displaystyle =\;\;-\dfrac{1}{x} + 1 - \dfrac{1}{x} + \dfrac{1}{x^2} \;\;=\;\;\dfrac{1}{x^2} - \dfrac{2}{x} + 1 $
Factor: . $\displaystyle \left(\dfrac{1}{x} - 1\right)^2$
$\displaystyle -x^{-1}+1-(x-1)x^{-2}$
$\displaystyle =-\frac{1}{x}-(-1)-(x-1)x^{-2}$
$\displaystyle =-\left(\frac{1}{x}-1\right)+(1-x)x^{-2}$
$\displaystyle =-\left(\frac{1}{x}-1\right)+\frac{1}{x}(1-x)x^{-2}x$
$\displaystyle =-\left(\frac{1}{x}-1\right)+\left(\frac{1}{x}-1\right)x^{-1}$
$\displaystyle =\left(\frac{1}{x}-1\right)\left(x^{-1}-1\right)=\left(\frac{1}{x}-1\right)\left(\frac{1}{x}-1\right)$