Question 1875
8 is {{{2^3}}} and 27 is {{{3^3}}} so, we can write the statement as:

{{{(2x)^3 + (3y)^3}}}

There is an identity that says

{{{(a^3 + b^3) = (a+b)(a^2+b^2-ab)}}} which is factorised completely,

Hence, our a is 2x and our b is 3y. Therefore we get

{{{(2x+3y)((2x)^2 + (3y)^2 - 6xy)}}}

cheers
jon.