Question

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Answer 1

multiply all the exponents by \(-2\) or else flip everything first then square it all

Answer 2

\[\large (-4a^2b^3)^{-2}=\frac{1}{(4a^2b^3)^2}\]

Answer 3

oops i dropped the minus sign
no matter since \((-4)^2=16\) anyways

Answer 4

final answer will be
\[\frac{1}{16a^4b^6}\]

Answer 5

Thank you, I think I am starting to understand how it works.. not fully yet though