# proving $a^{8} - 1 = \left(a^{2} - 1\right)\left(a^{2} + 1\right) \left(a^{2}+\sqrt{\,2\,}a+1\right) \left(a^{2} - \,\sqrt{\,2\,}\,a +1\right)$

• For all real numbers it is true that $$a^{8} - 1 = \left(a^{2} - 1\right)\left(a^{2} + 1\right) \left(a^{2} + \,\sqrt{\,2\,}\,a + 1\right) \left(a^{2} - \,\sqrt{\,2\,}\,a +1\right)$$ How do I arrive at this ?.
• I started thinking that $a^{8} -1$ is the $3$rd binomial formula so that $a^{8} - 1 = \left(a^{4} + 1\right)\left(a^{4} - 1\right)$, but then it basically says "stop" in my mental sphere.

$$(a^2)^2+1=(a^2+1)^2-(\sqrt2a)^2=\cdots$$