How do you simplify #root3(81x^8 y^12)#?

1 Answer

Answer:

#3^(4/3) x^(8/3) y^4=3x^2y^4root(3)(3x^2)#

Explanation:

With the expression #root(3)(81x^8y^12)#, we're taking the cube root of all the terms. We can write the cube root as a power of #1/3#, and so we get:

#(81 x^8 y^12)^(1/3)#

We can now distribute the #1/3# exponent to all the individual terms:

#81^(1/3) xx (x^8)^(1/3) xx (y^12)^(1/3)#

I'm going to express #81=3^4# and will also use the rule that #(x^a)^b=x^(ab)#:

#3^(4xx(1/3)) xx x^(8xx(1/3)) xx y^(12xx(1/3))#

which we can simplify:

#3^(4/3) xx x^(8/3) xx y^(12/3)#

#3^(4/3) xx x^(8/3) xx y^4#

We can now take the fractional exponents and where there is even divisibility, that lies outside of the root. Where there is a remainder, that can stay as a fraction or be shown under a root sign:

#3^(3/3) xx 3^(1/3) xx x^(6/3) xx x^(2/3) xx y^4#

#3x^2y^4 3^(1/3)x^(2/3)=3x^2y^4root(3)(3x^2)#