We can first work this and keep it in terms of exponents. Since #1=1^3 and 8=2^3#, we can write:

#(1/8)^(5/2)=(1^3/2^3)^(5/2)=((1/2)^3)^(5/2)=(1/2)^(3xx5/2)=(1/2)^(15/2)#

(this also uses the rule that #(x^a)^b=x^(ab)#)

~~~~~

We can also evaluate this numerically.

Let's first see that the exponent #15/2# means that we're going to take the fraction #1/2# and take the square root (that's the 2 in #15/2#) and also take #1/2# to the 15th power (that's the 15 in #15/2#).

Let's talk about the numerator first.

#1^(15/2)=1#

And now the denominator:

#(2)^(15/2)=2^(14/2+1/2)=2^(7+1/2)=2^7xx2^(1/2)=128sqrt2#

(we used the rule that #x^a xx x^b=x^(a+b)#)

We can then say that:

#(1/2)^(15/2)=1/(128sqrt2)#

We can then rationalize the denominator:

#1/(128sqrt2)(sqrt2/sqrt2)=sqrt2/(128xx2)=sqrt2/256#