Question 102440
:
{{{(x^3+8)/(x^2-2x+4)}}}
:
Note that the numerator can be factored as the "sum of cubes"
And one of the these factors is that same as the denominator, so they cancel

{{{((x+2)(x^2 - 2x + 4))/(x^2-2x+4)}}} = (x+2)