How do you simplify #( 9 / 49) ^ (- 3 / 2)#?

As per property:
#(a/b)^color(blue)(m)= a^color(blue)(m)/(b^color(blue)(m#

Applying the above to the expression :

#(9/49)^ (-3/2) = 9^color(blue)(-3/2)/(49^color(blue)(-3/2#

#(3^2)^(color(blue)(-3/2))/((7^2)^color(blue)(-3/2#

#=(3^cancel2)^(-3/cancel2)/((7^cancel2)^(-3/cancel2)#

#color(blue)("~~~~~~~~~~~~~~Tony B Formatting test~~~~~~~~~~~~~~~~~~~")#
#(3^(cancel(2))) (3/(cancel(2)))#

# (3^(cancel(2)))^(3/(cancel(2))) #

#color(red)("The formatting code can not cope with changing the second")# #color(red)("bracket group into index form.")#
#color(blue)("'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~")#

#=3^-3/(7^-3#

#=(1/27)/(1/343)#

#=343/27#

The minus in front of the index is instruction that this is a reciprocal

So we have: #1/((9/49)^(3/2))#

This is #((49)^(3/2))/((9)^(3/2))#

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Consider #color(white)(..)9^(3/2)#

This is the same as #(sqrt(9)color(white)(.) )^3=3^3=27 #

Giving: #((49)^(3/2))/27#

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Consider: #49^(3/2)#

This is the same as #(sqrt(49))^3=7^3=343 #

Giving:# (343)/27 = 12 19/27#