Question 127862
(3x+3y+3)/9x ÷ (x^2+ 2xy+y^2- 1)/(x^4+ x^2 ) =
[(3x+3y+3)/9x]  *   [(x^4+ x^2 )/(x^2+ 2xy+y^2- 1)] =
(3x+3y+3)*(x^4+ x^2 )/9x*(x^2+ 2xy+y^2- 1)=
(3x+3y+3)(x^3+x)/(9x^2+18xy+9y^2-9)=
(x+y+1)*(x^3+x)/(3x^2+6xy+3y^2-3)