How do I factor 64x^3-27y^12

This also looks like the difference of two cubes (a^{3} - b^{3}) = (a - b) (a^{2} + ab + b^{2}), since

64 = 4^{3} and 27 = 3^{3}. Also, y^{12} = (y^{4})^{3} So,

64x^{3} - 27y^{12} = (4x)^{3} - (3y^{4})^{3}

(4x - 3y^{4})[(4x)^{2} + (4x)(3y^{4}) + (3y^{4})^{2}]

(4x - 3y^{4})[16x^{2} + 12xy^{4} + 9y^{8}]