Question 175719
I believe what you are asking is to factor
{{{8x^3y^6 + 27}}}
Notice that each term can be written using cubes.
{{{8x^3y^6 + 27 = (2^3)(x^3)(y^2)^3 + 3^3}}}
So we have that
{{{8x^3y^6 + 27 = (2xy^2)^3 + 3^3}}}
Now, the sum of cubes can be factored as following:
{{{a^3+b^3=(a+b)(a^2-ab+b^2)}}}
So, here {{{a=2xy^2}}} and b=3 
{{{8x^3y^6 + 27 = (2xy^2)^3 + 3^3 = (2xy^2 + 3)((2xy^2)^2 -2xy^2(3) + 3^2)}}}
Simplified, this is
{{{(2xy^2 + 3)(4x^2y^4 - 6xy^2 + 9)}}}