Question 404577
{{{x^8 - 1}}}..........{{{8=2*2*2}}}, so you can write {{{((x^4)^2 )))}}} or 

{{{((x^2)^2)^2 )))}}} ...same with {{{1}}}


{{{(x^4)^2 -(1^4)^2}}}.....use a rule for factoring a difference of squares


{{{(x^4 -1^4)(x^4+ 1^4)}}} ..............{{{x^4= (x^2)^2)}}}

and we have

{{{(((x^2)^2) -((1^2)^2))(x^4+ 1)}}}..factor this {{{(((x^2)^2) -((1^2)^2))}}}


{{{(x^2 -1)(x^2 +1)(x^4+ 1)........factor {{{(x^2 -1)}}}


{{{(x -1)(x +1)(x^2 +1)(x^4+ 1)}}}


and 


{{{x^4 + 2x^3 - 3x - 6}}}...group  


{{{(x^4 + 2x^3) - (3x + 6)}}}


{{{x^3(x + 2) - 3(x + 2)}}}.....common factor {{{(x + 2)}}}


{{{(x + 2)(x^3- 3)}}}