simplifying the negative exponent

[spacewater]

The problem
\(\displaystyle X^5 - 2x^{-2}\)
Steps I've taken
\(\displaystyle \frac{2x^{2}}{2x^{2}} \cdot \frac{x^5}{1} - \frac {1}{2x^{2}}{\)
to make the denominator equal
\(\displaystyle \frac {2x^{7}}{2x^2} - \frac {1}{2x^2} = \frac {2x^{7} -1}{2x^2}\)
canceled out 2 in the numerator and 2 in the denominator to get
\(\displaystyle \frac {x^{7}-1}{x^2}\)

But the answer-sheet says its wrong and I can't seem to figure out where I went wrong.
Can anybody who isnt too busy elaborate this problem for me please? :roll:

 

[Loren]

\(\displaystyle \frac{2a-4}{2}=\frac{\rlap{/}2a-4}{\rlap{/}2}=\frac{a-4}{1}\). <<<WRONG!
\(\displaystyle \frac{2a-4}{2}=\frac{\rlap{/}2(a-2)}{\rlap{/}2}= \frac{a-2}{1}\) <<<CORRECT!

You cannot cancel terms. You can cancel factors. Now. Be sure you know the difference between a term and a factor.

 

[Deleted member 4993]

The problem
\(\displaystyle X^5 - 2x^{-2}\) .................why do you have capital X and small x...I assume those are same

\(\displaystyle = \, x^5 \, - \, \frac{2}{x^2}\)

\(\displaystyle = \, \frac {x^5\cdot x^2}{x^2} \, - \, \frac{2}{x^2}\)

Now continue....

________________________________________________________
Steps I've taken
\(\displaystyle \frac{2x^{2}}{2x^{2}} \cdot \frac{x^5}{1} - \frac {1}{2x^{2}}{\)
to make the denominator equal
\(\displaystyle \frac {2x^{7}}{2x^2} - \frac {1}{2x^2} = \frac {2x^{7} -1}{2x^2}\)
canceled out 2 in the numerator and 2 in the denominator to get
\(\displaystyle \frac {x^{7}-1}{x^2}\)

But the answer-sheet says its wrong and I can't seem to figure out where I went wrong.
Can anybody who isnt too busy elaborate this problem for me please? :roll: