Answer
$(5x^2-4y^2)(25x^4+20x^2y^2+16y^4)$
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: $125x^6-64y^6=\\=(5x^2)^3-(4y^2)^3\\=(5x^2-4y^2)((5x^2)^2+5x^2\cdot4y^2+(4y^2)^2)\\=(5x^2-4y^2)(25x^4+20x^2y^2+16y^4)$