The expression is equal to:
[tex](\frac{-3jk^3}{2k^2m^3})^3=(\frac{-3jk^{(3-2)}}{2m^3})^3=(\frac{-3jk}{2m^3})^3[/tex][tex](\frac{-3jk^3}{2k^2m^3})^3=\frac{(-3)^3\cdot j^3\cdot k^3}{2^3\cdot(m^3)^3}=\frac{-27j^3k^3}{8m^{(3\cdot3)}}=\frac{-27j^3k^3}{8m^9}[/tex]
So:
[tex]\frac{Aj^xk^y}{Bm^z}=\frac{-27j^3k^3}{8m^9}[/tex]
Answer: A = -27
B = 8
x = 3
y = 3
z = 9
