Question 600727
{{{a^8-b^8}}}


{{{(a^4)^2-(b^4)^2}}}


{{{(a^4-b^4)(a^4+b^4)}}}


{{{((a^2)^2-(b^2)^2)(a^4+b^4)}}}


{{{(a^2-b^2)(a^2+b^2)(a^4+b^4)}}}


{{{(a-b)(a+b)(a^2+b^2)(a^4+b^4)}}}


So {{{a^8-b^8}}} completely factors to {{{(a-b)(a+b)(a^2+b^2)(a^4+b^4)}}}