Question 382946
First factor.
{{{(x^3+8)/(x^2-4)=((x+2)(x^2-2x+4))/((x+2)(x-2))}}}
{{{(x^3+8)/(x^2-4)=(x^2-2x+4)/(x-2)}}}
.
.
{{{(x^2-4x+4)/(x^2-2x+4)=(x-2)^2/(x^2-2x+4)}}}
So then,
{{{((x^3+8)/(x^2-4))*((x^2-4x+4)/(x^2-2x+4)) =((x^2-2x+4)/(x-2))*((x-2)^2/(x^2-2x+4))}}}
{{{((x^3+8)/(x^2-4))*((x^2-4x+4)/(x^2-2x+4)) =highlight((x-2))}}}