Question 373942
{{{(x^2-4x+16)/(x^3+64)}}}
There is a pattern for sum of cubes: {{{a^3+b^3 = (a+b)(a^2-ab+b^2)}}}. Since {{{64 = 4^3}}} the denominator fits this pattern. SO we can use the pattern to factor the denominator. With the numerators all we can do is factor out 1:
{{{(1*(x^2-4x+16))/((x+4)((x)^2 - (x)(4) + (4)^2))}}}
which simplifies as follows:
{{{(1*(x^2-4x+16))/((x+4)(x^2 - 4x + 16))}}}
{{{(1*cross((x^2-4x+16)))/((x+4)cross((x^2 - 4x + 16)))}}}
{{{1/(x+4)}}}