Question 946254
I don't understand how (n-2)/(n^3-8) becomes 1/(n^2+2n+4).

I get how you take out the common factor of (n-2), but why does the denominator become n^2+2n+4 instead of n^2+4?
<pre>
{{{(n - 2)/(n^3 - 8)}}}
{{{(n - 2)/((n - 2)(n^2 + 2n + 4))}}} ----- Factoring denominator
{{{cross((n - 2))/(cross((n - 2))(n^2 + 2n + 4))}}} ---- Cancelling n - 2 in numerator and denominator
{{{highlight_green(1/(n^2 + 2n + 4))}}}