Question 994474


{{{a^6-b^6}}}

{{{(a^3)^2-(b^3)^2}}}

{{{(a^3-b^3)(a^3+b^3)}}}.....use the rule for factoring {{{(a^3-b^3)}}} and {{{(a^3+b^3)}}}:{{{a^3-b^3=(a-b)(a^2-ab+b^2)}}} and {{{a^3+b^3=(a+b)(a^2+ab+b^2)}}}

so, we have:

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