+770 Let's break this up :D
There's \(3a^{-3}\), \(b^2\), \(2a^{-1}\) and \(b^0\), where \(a = -2\) and \(b=-3\).
\(3a^{-3} = 3((-2)^{-3}) = 3(\frac{1}{(-2)^3}) = \frac{3}{-8} = -\frac{3}{8}\)
\(b^2 = (-3)^2 = 9\)
\(2a^{-1} = 2((-2)^{-1}) = 2(\frac{1}{-2}) = \frac{2}{-2} = -1\)
\(b^0 = 1 \) :D
So, our expression is:
\((\dfrac{-\frac{3}{8} \cdot 9}{-1 \cdot 1})^2\)
Which is:
\((\dfrac{-\frac{27}{8}}{-1})^2\)
Which makes:
\((\frac{27}{8})^2\)
Which equals:
\(\fbox{$\frac{729}{64}$}\)
Which is the last option :D
