Question 607215
{{{(y^3+64)/(y^3-4y+16)}}} Factor the numerator (the sum of two cubes: {{{(y)^3+(4)^3}}}).
{{{(y+4)(y^2-4y+16)/(y^3-4y^2+16y)}}} Factor the denominator.
{{{(y+4)cross((y^2-4y+16))/(y*cross((y^2-4y+16)))}}} Cancel the common factors as indicated to get:
{{{(y+4)/y}}}