<s>[INST] How to show that $\int_{0}^{\infty}{(2x)^4\over (1-x^2+x^4)^4}\mathrm dx=\int_{0}^{\infty}{(2x)^4\over (1-x^2+x^4)^3}\mathrm dx=3\pi$ 
Consider
  $$I=\int_{0}^{\infty}{(2x)^4\over (1-x^2+x^4)^4}\mathrm dx
\qquad J=\int_{0}^{\infty}{(2x)^4\over (1-x^2+x^4)^3}\mathrm dx$$
  I want to show that $I=3\pi$ and that $I=J$.

First, we noticed that $x^4+x^2+1=(x^2-x+1)(x^2+x+1)$
So it gives us an idea to try and factorise $x^4-x^2+1$ but cannot find any factors.
Integrate $I$ (We try some substitutions to see where it will get us to)
$x=\sqrt{u}$ then $dx={2\over \sqrt{u}}du$
$$I=16\cdot{1\over 2}\int_{0}^{\infty}{u^{3/2}\over (1-u+u^2)^4}\mathrm du$$
$u=\tan(y)$ then $du=\sec^2(y)dy$
$$I=8\int_{0}^{\pi/2}{\tan^{3/2}(y)\over (1-\tan(y)+\tan^2(y))^4}{\mathrm dy\over \cos^2(y)}$$
then simplified down to
$$I=128\int_{0}^{\pi/2}{\cos^6(y)\tan^{3/2}(y)\over (2-\sin(2y))^4}\mathrm dy$$
we further simplified down to
$$I={128\over 2^{3/2}}\int_{0}^{\pi/2}{\cos^3(y)\sin^{3/2}(2y)\over (2-\sin(2y))^4}\mathrm dy$$
Not so sure what is the next step.
 [/INST] Consider the transformation $x=1/y$. Then
$$
J=16\int_0^{\infty} dy\frac{1}{y^2} \frac{1/y^4}{(1-1/y^2+1/y^4)^3}=8\int_{-\infty}^{\infty} dy\frac{1}{(y^2-1+1/y^2)^3}=\\8\int_{-\infty}^{\infty} dy\frac{1}{((y-1/y)^2+1)^3}\underbrace{=}_{(\star)}8\int_{-\infty}^{\infty}dz\frac{1}{(z^2+1)^3}=3\pi
$$
and your proof is complete (for the next to last equality $(\star)$ we applied Glasser's Master theorem which is not difficult to proof for this special case)
Furthermore applying the same substitution $x=1/y$ again, it is a matter of straightforward algebra that
$$
\Delta=J-I=-\Delta
$$
so 

$$
\Delta=0 \,\,\text{or}\,\,I=J=3\pi
$$

QED
</s>