Question 252895
A special case of factoring is "sum of cubes":
{{{ a^3 + b^3 = (a + b)(a^2 - ab + b^2) }}}
.
Your problem:
{{{a^6+b^6}}}
can be rewritten as:
{{{(a^2)^3+(b^2)^3}}}
which is "sum of cubes" which we now factor as:
{{{(a^2+b^2)((a^2)^2-a^2b^2+(b^2)^2)}}}
{{{(a^2+b^2)(a^4-a^2b^2+b^4)}}}