How do you simplify #(root4(5))/(4root4(27))#?

1 Answer
Dec 19, 2016

Answer:

#(root(4)15)/12#

Explanation:

#(root(4)5)/(4*root(4)27)#, we first want to remove the radicals from the denominator, we can do this by multiplying by #root(3/4)27/root(3/4)27# which is equal to #1#.

#(root(4)5)/(4*root(4)27) * root(4)27^3/root(4)(27)^3 = (root(4)5*root(4)27^3)/(4*root(4)27*root(4)27^3) = (root(4)5*root(4)27^3)/(4*root(4)27^4) = (root(4)5*root(4)27^3)/(4*27)#

#=(root(4)5*root(4)27*root(4)27*root(4)27)/108 = (root(4)(5*27^3))/108 =(root(4)(5*3^9))/108 = (root(4)(5*3^4 * 3^4 * 3))/108 =(3*3root(4)(5*3))/108 = (9root(4)(15))/(9*12) = (cancel9root(4)(15))/(cancel9*12)#

This is as simplified as it gets,

#(root(4)15)/12#