First, recall that anything to the power of #0# is equal to #1# (with the exception of #0^0.#) So, all instances of #y^0# are replaced with #1#.
#(xy^2)^4/(2x^-3)^2#
Recall that #(xy)^a=x^ay^a#. That is, whenever we have a group of factors raised to a power, we multiply the power through.
#(xy^2)^4= x^4y^((2)(4))=x^4y^8#
Do the same for the bottom:
#(2x^-3)^2=2^2x^((-3)(2))=4x^-6#
Let's rewrite our expression:
#(x^4y^8)/(4x^-6)#
Recall that #x^a/x^b=x^(a-b)#. That is, whenever we divide two factors with the same base that are both raised to certain powers, we subtract the power of the denominator from the power of the numerator. Here, this is the case with the #(x^4/x^-6)#.
Our expression becomes:
#(x^(4-(-6)y^8)/(4))=((x^10y^8)/4)#
