Question 360217
Assume the problem is:
{{{((w^2-y^2) * (w^2+wy+y^2))/((w^3-y^3)* (w^2+2wy+y^2))}}} 
We have to factor here, we have the difference of squares  the difference of cubes
{{{((w-y)*(x+y) * (w^2+wy+y^2))/((w-y)*(w^2+wy+y^2)* (w^2+2wy+y^2))}}}
We can also factor (w^2+2wy+y^2)
{{{((w-y)*(x+y) * (w^2+wy+y^2))/((w-y)*(w^2+wy+y^2)* ((w+y)*(w+y)))}}}
Look at all we can cancel here: (w-y), (w+y) and (w^2+wy+y^2), that gives us:
{{{1/((w+y))}}}