The answer is B.
To help prove it you can do a multiples tree. Multiply all the like terms together and you get the negative cubed root of 40x^9 times y^3. Now you just look for pairs of three to put on the outside of the cubed root. 40 = 2 x 2 x 2 x 5 so the triple 2 turns into one 2 on the outside of the cube root. There are nine X's which turns into three x on the outside of the root. There are three y's which turn into one y on the outside of the root. There is still a negative on the outside so it is negative 6x^3y times the cubed root of 5.