Not sure if my workbook is wrong or my calculation. Probably me :)
Here is the problem:
$\frac {k^3-1}{k-1}$
My work: (inserted some zeros because of my lack of MathJax skills)
$ \begin{array}{r} \text{} \underline{k^2+k+1} \\ k-1 | k3 + 0 +0-1 \\ \underline{k^3-k^2+0-1\phantom{0}} \\ k^2+0-1 \\ \underline{k^2-k+0\phantom{0}} \\ k-1 \\ \underline{k-1\phantom{0}} \\ 0 \end{array} $
My answer is: $k^2+k+1$
My workbook gives the answer of: $k^2+k+1+\frac{2}{k-1}$
Which is correct?
Your answer is correct.
Remark:
This is a special case of a famous formula and it is known as geometric series.
In general $$1+k +\ldots + k^{n-1} = \frac{k^n-1}{k-1}$$
For your case $n=3$.