Dividing by a Binomial Denominator

Quadratic equations and inequalities, variation equations, function notation, systems of equations, etc.
RedComet
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Dividing by a Binomial Denominator

I think I've found a mistake in the review book I'm using. Would someone care to work this problem and see if they get the same thing?

(k^3 -1) / (k - 1)

I ended up with k^2 + k + 1 as the answer, but the book gives the answer as k^2 - k - 1 - (2 / [k - 1])

Thanks for the help.

P.S. Awesome site. I've been using this since high school (5+ years ago) and I'm using it again in conjunction with other books to get myself ready for college level Calculus.

nona.m.nona
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Re: Dividing by a Binomial Denominator

(k^3 -1) / (k - 1)

I ended up with k^2 + k + 1 as the answer...
This result can be obtained by factoring the difference of cubes and then canceling.
...but the book gives the answer as k^2 - k - 1 - (2 / [k - 1])
Simplify:

$\left(\frac{k^2\, -\, k\, -\, 1}{1}\right)\left(\frac{k\, -\, 1}{k\, -\, 1}\right)\, -\, \frac{2}{k\, -\, 1}$

$\frac{k^3\, -\, 2k^2\, +\, 1}{k\, -\, 1}\, -\, \frac{2}{k\, -\, 1}$

$\frac{k^3\, -\, 2k^2\, -\, 1}{k\, -\, 1}$

This does not equal the original expression, so the simplification would appear to be incorrect.

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