# Partial Fractions Solving: Denominators having degrees more than 2

 P: 6 I just want to know if there is now a way to solve fractions like which had a variables that has a degree more than 2 in its denominator. I know that denominators having degrees of 2 could be solved using (Ax + B)/(x2+a). But how about denominators like (x3+a) and so on???
 Sci Advisor P: 6,039 The main difficulty is that it is much harder to get the factors for higher degree polynomials. The principal is the same once you've got the factors.
 Math Emeritus Sci Advisor Thanks PF Gold P: 39,345 On the other hand, the one you specifically give is relatively simple: As checkitagain pointed out, this should have been $$x^2+ y^2= (x+ y)(x^2- xy+ y^2)$$ By "completing the square", $x^2- xy+ y^2= x^2- xy+ y^2/4- y^2/4+ y^2= (x- y/2)^2+ 3y^2/4$. Let $y= \sqrt{a}$. But, as mathman says, while every polynomial can be factored into linear or quadratic terms over the real numbers (into linear terms over the complex numbers), the higher the degree of the polynomial, the harder to find the factors.
P: 99
Partial Fractions Solving: Denominators having degrees more than 2

 Quote by HallsofIvy On the other hand, the one you specifically give is relatively simple: $$x^2+ y^2= (x+ y)(x^2+ xy+ y^2)$$
in the quote box above, so I am posting this:

$$x^3 - y^3 = (x - y)(x^2 + xy + y^2)$$

$$x^3 + y^3 = (x + y)(x^2 - xy + y^2)$$
 P: 6 so it can only be done to perfect cube terms??
 Quote by checkitagain I haven't received any feedback about my issues with the content in the quote box above, so I am posting this: $$x^3 - y^3 = (x - y)(x^2 + xy + y^2)$$ $$x^3 + y^3 = (x + y)(x^2 - xy + y^2)$$