Given:
#" "#
#color(blue)(16^(-2/3)#
Identities used:
#color(red)(a^(-b)=1/(a^b)#
#color(red)(a^(m/n) = rootn(a^m)#
Consider the expression given:
#color(blue)(16^(-2/3)#
#rArr 1/(16^(2/3))#
#rArr 1/root3(16^2)#
#rArr 1/root3(256#
#rArr 1/root3(64*4)#
Note that #color(blue)(64 = 4^3#
#rArr 1/root3(4^3*4)#
#rArr 1/root3(4^3)*1/root3(4)#
Note that #color(red)(rootp(m^n) = (m^n)^(1/p)#
#rArr 1/(4^3)^(1/3)*1/root3(4)#
#rArr (1/4)*1/root3(4)#
#rArr (1/4)*1/root3(2^2)#
Note that #color(red)(sqrt(m^n) = (m^n)^(1/2)=m^(n/2)#
#rArr 1/4*[1/[(2^2)^(1/3))]#
#rArr 1/4*1/(2^(2/3))#
#rArr 1/(4*2^(2/3)#
Hence,
#color(blue)(16^(-2/3) = 1/(4*2^(2/3)#
If you wish, you can continue to simplify further:
#1/(4*2^(2/3)#
#rArr 1/(2^2*2^(2/3)#
Note that #color(red)(a^m*a^n=a^(m+n)#
#rArr 1/2^(2+(2/3)#
#rArr 1/(2^(8/3)#
#rArr 2^(-8/3#
Hope you find this solution useful.
