Question 169991
{{{(a^(2n))-1}}} Start with the given expression



{{{(a^n)^2-1}}} Rewrite {{{a^(2n)}}} as {{{(a^n)^2}}}



{{{(a^n)^2-1}}} Rewrite {{{1}}} as {{{(1)^2}}}



Notice we have a difference of squares {{{A^2-B^2}}}. In this case, {{{A=a^n}}} and {{{B=1}}}



So to factor this difference of squares, we can use the formula {{{A^2-B^2=(A+B)(A-B)}}}



So this means that {{{(a^n)^2-1=(a^n+1)(a^n-1)}}}



So {{{(a^n)^2-1}}} factors to {{{(a^n+1)(a^n-1)}}}