How to factor 64x^4 + xy^3

So we kkonw that 64=4^3

so if we factor out x we get

x (4^3 (x^3) + y^3) or x[ (4x) ^3+y^3)

so the additon of two perfect cubes so

(4x+y) (16x^2+4xy+y^2)

and we remember that this is all times x so

x[ (4x+y (16^2+4xy+y^2)) ]

or (x) (4x+y) (16x^2+4xy+y^2)

You factor out an x and get

x (64x^3+y^3)