Question 607499
Apply the special factoring case of "difference of cubes":
{{{a^3-b^3 = (a-b)(a^2+ab+b^2)}}}
.
Looking at the numerator:
{{{a^3-1}}}
{{{a^3-1^3}}}
{{{(a-1)(a^2+a(1)+1^2)}}}
{{{(a-1)(a^2+a+1)}}}
.
so, now we can rewrite:
{{{(a^3-1)/(a-1) }}}
as:
{{{((a-1)(a^2+a+1))/(a-1) }}}
canceling like-terms:
{{{a^2+a+1}}}