Answer:
[tex]\frac{729}{64}[/tex]
Step-by-step explanation:
The given expression is
[tex](\frac{3a^{-3}b^{2} }{2a^{-1} b^{0} } )^{2}[/tex]
Where [tex]a=-2[/tex] and [tex]b=-3[/tex].
When we say "evaluate the expression" refers to replacing the given values in the expression. Then, we solve each power in the numerator and denominator. After that, we solve each product in the fraction. At the end, we apply the square power. As follows
[tex](\frac{3a^{-3}b^{2} }{2a^{-1} b^{0} } )^{2}=(\frac{3(-2)^{-3}(-3)^{2} }{2(-2)^{-1} (-3)^{0} } )^{2}=(\frac{3(\frac{1}{-8})(9)}{2(\frac{1}{-2} (1))} )^{2} =(\frac{-\frac{27}{8} }{-1} )^{2} =\frac{27^{2} }{8^{2} } =\frac{729}{64}[/tex]
