Resolve $$x^8 + y^8$$ into real quadratic factors.

PLACEHOLDER ANSWER - the answer is a product of four quadratics. The roots of $$t^8 = -1$$ are $$e^{i(2k + 1)\pi/8}$$, and pairing conjugates gives $$x^8 + y^8 = \prod_{k=0}^{3}\left(x^2 - 2xy\cos\frac{(2k + 1)\pi}{8} + y^2\right)$$. Since $$2\cos\frac{\pi}{8} = \sqrt{2 + \sqrt{2}}$$ and $$2\cos\frac{3\pi}{8} = \sqrt{2 - \sqrt{2}}$$, this is $$\left(x^2 + \sqrt{2 + \sqrt{2}}\,xy + y^2\right)\left(x^2 - \sqrt{2 + \sqrt{2}}\,xy + y^2\right)\left(x^2 + \sqrt{2 - \sqrt{2}}\,xy + y^2\right)\left(x^2 - \sqrt{2 - \sqrt{2}}\,xy + y^2\right)$$.