Question 150362
First let's simplify.
Let A={{{x^3+3x^2-x-3}}}, then
{{{3/(x^3+3x^2-x-3)-2/(x^3-x^2-3x+3)+4/(x^3+x^2-3x-3)=3/A-2/A+4/A}}}
Use the distributive property.
{{{3/(x^3+3x^2-x-3)-2/(x^3-x^2-3x+3)+4/(x^3+x^2-3x-3)=3/A-2/A+4/A}}}
{{{3/(x^3+3x^2-x-3)-2/(x^3-x^2-3x+3)+4/(x^3+x^2-3x-3)=(1/A)(3-2+4)}}}
{{{3/(x^3+3x^2-x-3)-2/(x^3-x^2-3x+3)+4/(x^3+x^2-3x-3)=(1/A)(5)}}}
{{{3/(x^3+3x^2-x-3)-2/(x^3-x^2-3x+3)+4/(x^3+x^2-3x-3)=5/(x^3+3x^2-x-3)}}}