What is the correct radical form of this expression #(32a^10b^(5/2))^(2/5)#?

1 Answer
mason m
Dec 12, 2016

#(32a^10b^(5/2))^(2/5)=4a^4b#

Explanation:

First, rewrite #32# as #2xx2xx2xx2xx2=2^5#:

#(32a^10b^(5/2))^(2/5)=(2^5a^10b^(5/2))^(2/5)#

The exponent can be split up by multiplication, that is, #(ab)^c=a^c*b^c#. This is true for a product of three parts, such as #(abc)^d=a^d*b^d*c^d#. Thus:

#(2^5a^10b^(5/2))^(2/5)=(2^5)^(2/5) * (a^10)^(2/5)*(b^(5/2))^(2/5)#

Each of these can be simplified using the rule #(a^b)^c=a^(bc)#.

#(2^5)^(2/5) * (a^10)^(2/5) * (b^(5/2))^(2/5)=2^(5xx2/5) * a^(10xx2/5) * b^(5/2xx2/5)#

#color(white)((2^5)^(2/5) * (a^10)^(2/5) * (b^(5/2))^(2/5))=2^2 * a^4 * b^1#

#color(white)((2^5)^(2/5) * (a^10)^(2/5) * (b^(5/2))^(2/5))=4a^4b#

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