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) $

Question

• 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.

I'm preparing for an admissions exam please help !. THX.

Answer 1

\begin{align} (a^8-1) &= (a^4-1)(a^4+1) \\ &= (a^2-1)(a^2+1)(a^4+2a^2+1 - 2a^2) \\ &= (a^2-1)(a^2+1)((a^2+1)^2 - 2a^2) \\ &= (a^2-1)(a^2+1)((a^2+1) - \sqrt{2}a)((a^2+1) + \sqrt{2}a) \end{align}

Answer 2

HINT:

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