Factoring irreducible fractions Is there a technique or a formula that factors irreducible fractions into products of polynomials, for example:
$$\frac1{x^8+1}=\frac{1}{(x^4+ \sqrt2x^2+1)(x^4- \sqrt2x^2+1)}$$
Also, is there a way to factor, in the same way as above, the following fraction:
$$\frac1{x^4-x^2+1}$$
I am not interested in the completition of the square. Thank you.
 A: You can factor second fraction as $$\frac { 1 }{ x^{ 4 }-x^{ 2 }+1 } =\frac { 1 }{ x^{ 4 }+2x^{ 2 }+1-3{ x }^{ 2 } } =\frac { 1 }{ { \left( { x }^{ 2 }+1 \right)  }^{ 2 }-{ \left( \sqrt { 3 } x \right)  }^{ 2 } } =\frac { 1 }{ \left( { x }^{ 2 }+1-\sqrt { 3 } x \right) \left( { x }^{ 2 }+1+\sqrt { 3 } x \right)  }  $$
A: One has a direct way, writing
$$
x^8+1=(x^8+2x^4+1)-2x^4=(x^4+1)^2-2x^4=(x^4+ \sqrt2x^2+1)(x^4- \sqrt2x^2+1)
$$ similarly
$$
x^4-x^2+1=(x^4+1)-x^2=(x^2+1)^2-3x^2=(x^2+ \sqrt3x+1)(x^2- \sqrt3x+1).
$$
A: To answer the "without completion of the square" part of the question, it is possible to actually determine the roots of the given polynomials in both cases. Then grouping each root with its complex conjugate gives a quadratic in $x$ which is a factor of the original polynomial.
In the case of $x^8+1$ for example, a pair of roots is $x = \cos(\frac{\pi}{8}) \pm i \sin(\frac{\pi}{8})\,$. The quadratic which has those roots is $x^2 - 2 \cos(\frac{\pi}{8})\,x +1\,$. Using the half-angle formula, $\cos(\frac{\pi}{8})= \frac{1}{2}\sqrt{2+\sqrt{2}}\,$. Therefore a factor of $x^8+1$ is:
$$x^2 - \sqrt{2+\sqrt{2}} \,x + 1$$
Since $x^8 +1$ is an even polynomial, another factor follows by substituting $x \mapsto -x\,$:
$$x^2 + \sqrt{2+\sqrt{2}} \,x + 1$$
Their product, which itself must be a factor of $x^8+1$ is:
$$\left(x^2 - \sqrt{2+\sqrt{2}} \,x + 1\right)\left(x^2 + \sqrt{2+\sqrt{2}} \,x + 1\right) \;=\; x^4 - \sqrt{2} \,x^2 + 1$$
Dividing $x^8+1$ by the above gives the other factor $\;x^4 + \sqrt{2} \,x^2 + 1\,$.
