Question 57356
The formula for factoring the sum of perfect cubes is: {{{u^3+v^3=(u+v)(u^2-uv+v^2)}}}.  We'll need that to factor {{{x^3+64}}}, u=x and v=4
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Let's get started.
:
{{{(x^3+ 2x^2)/(x^3+64)}}} divided by {{{4x^2/(x^2-4x+16)}}} First flip and multiply:
{{{((x^3+2x^2)/(x^3+64))*((x^2-4x+16)/4x^2)}}} Factor everything.
{{{x^2(x+2)/((x+4)(x^2-4x+16))*((x^2-4x+16)/4x^2)}}}  Cancel the things that match between the numerators and the denominators.
{{{(cross(x^2)(x+2)/((x+4)*cross((x^2-4x+16))))*(cross((x^2-4x+16))/(4*cross(x^2)))}}}
{{{((x+2)/(x+4))*(1/4)}}}  Multiply what's left:
{{{(x+2)/(4(x+4))}}}  Most teachers let you stop here, but you can distribute the 4 and get:
{{{(x+2)/(4x+16)}}}
Happy Calculating!!!