How do you write #32^(-3/5)# in radical form?

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Answer 1 · 2018-06-04T08:56:29

Se explanation below

We define #a^(m/n)=root(n)(a^m)#.

By other hand, we know that #a^(-n)=1/a^n#

With these rules in mind, in our case

#32^(-3/5)=1/(root(5)(32^3))=1/(root(5)((2^5)^3))=1/(root(5)(2^15))=1/root(5)(2^5·2^5·2^5)=1/8#

Answer 2 · 2018-06-04T15:47:13

#1/8#

#32^(-3/5)#

#:.=(2^5)^(-3/5)#

#:.=2^(-15/5)#

#:.=2^-3#

#:.=m^-3=1/m^3#

#:.=1/2^3#

#:.=1/8#