Question 648536
Think of {{{(4g)^-3  = (4g)^(0-3)}}}

Since {{{n^(a-b) =  n^a / n^b}}}, then:

{{{(4g)^(-3) = (4g)^(0-3) = (4g)^0/(4g)^3 = 1/(4g)^3}}}

In general,  {{{ n^(-a) = 1/n^a.}}}

We can simplify the denominator

{{{1/(4g)^3 =  1/(4^3 *g^3) =  1/64g^3}}}