There are essentially three rules in these sorts of power multiplication problems:
[tex] x^{-a} = \dfrac{1}{x^a} [/tex]
[tex] x^a x^b = x^{a+ b} [/tex]
[tex] (x^a)^b = x^{a b} [/tex]
These rules just tell us a negative exponent means the reciprocal of the same thing with a positive exponent, that when we multiply factors with the same base we add their exponents, and when we raise something to a power to another power we get to multiply the exponents.
Let's do the algebra before we substitute (we could do it either way).
[tex] (p^2 q^{-3} )^{-2} (p^{-3}q^5)^{-2} [/tex]
[tex] = (p^{-4} q^{6} ) (p^{6}q^{-10}) [/tex]
[tex] = p^{2} q^{-4} [/tex]
Substituting p=-2, q=-1
[tex] = 2^{2} (-1)^{-4} [/tex]
[tex] = 4 (1/(-1)^4) = 4\ (1/1) = 4[/tex]
Answer: 4, last choice
