How do you simplify #(root3(9)*root3(6))/(root6(2)*root6(2))#?

1 Answer
Carlo Maximo
May 11, 2017

3.

Explanation:

#root(3)9 = 3^(1/3)*3^(1/3)#
#root(3)6 = 2^(1/3)*3^(1/3)#
#root(6)2 = 2^(1/6)#

Now you have #(3^(1/3)*3^(1/3)*2^(1/3)*3^(1/3))/(2^(1/6)*2^(1/6))#, right?
By Laws of Exponents, you could combine those #3^(1/3)# and #2^(1/6)# and add their bases.

Now you have #3*2^(1/3)/2^(1/3)#. Cancelling the fraction part, you'll get 3. :)

Related questions
See all questions in Multiplication and Division of Radicals
Impact of this question
1101 views around the world
You can reuse this answer
Creative Commons License