Answer
$(x^2+3y)(x^4-3x^2y+9y^2)$
Work Step by Step
The formula for factoring the sum of two cubes is: $A^3+B^3=(A+B)(A^2-AB+B^2)$.
The formula for factoring the difference of two cubes is: $A^3-B^3=(A-B)(A^2+AB+B^2)$.
Hence here: $x^6+27y^3=\\=(x^2)^3+(3y)^3\\=(x^2+3y)((x^2)^2-x^2\cdot3y+(3y)^2)\\=(x^2+3y)(x^4-3x^2y+9y^2)$
