Question 152956

{{{x^8-1}}} Start with the given expression



{{{(x^4)^2-(1)^2}}} Rewrite {{{x^8}}} as {{{(x^4)^2}}}. Rewrite {{{1}}} as {{{(1)^2}}}



{{{(x^4+1)(x^4-1)}}} Factor using the difference of squares formula. Remember, the difference of squares formula is {{{A^2-B^2=(A+B)(A-B)}}}



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



{{{(x^4+1)(x^2+1)(x^2-1)}}} Factor {{{(x^2)^2-(1)^2}}} using the difference of squares formula



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



{{{(x^4+1)(x^2+1)(x+1)(x-1)}}}   Factor {{{x^2-(1)^2}}} using the difference of squares formula 



So the expression {{{x^8-1}}} factors to {{{(x^4+1)(x^2+1)(x+1)(x-1)}}}



In other words, {{{x^8-1=(x^4+1)(x^2+1)(x+1)(x-1)}}}