Simplify #(root3(24)-root3(81))/(sqrt32xxsqrt(2xxroot3(9))#?

1 Answer

Answer:

#-1/8#

Explanation:

Let's work this by breaking down the different terms and see where we end up.

We start with:

#(root3(24)-root3(81))/(sqrt32xxsqrt(2xxroot3(9))#

I'll work through the cube roots in the numerator and the roots in the denominator:

First thing I'll do is break down the numbers within the roots to find the cube/square inside the root sign, which will allow us to simplify them. I'm also going to distribute the square root across the #sqrt(2root3(9))# to make that easier to work with.

#(root3(8xx3)-root3(27xx3))/(sqrt(16xx2)xxsqrt(2xxroot3(9))#

#(root3(2^3xx3)-root3(3^3xx3))/(sqrt(4^2xx2)xxsqrt(2)xxsqrt(root3(9))#

#(2root3(3)-3root3(3))/(4sqrt(2)xxsqrt(2)xxroot6(3^2))#

The numerator is now ready for a subtraction and have a single term up top. I can regroup the denominator to perform multiplication.

I'm also going to evaluate the #root6(3^2)# term by doing the following (I'll work this side bit in red - and I'll start it from the start of the question so you see the entire progression of work):

#color(red)sqrt((root3(9)))=color(red)((9^(1/3))^(1/2))=color(red)(9^(1/3xx1/2))=color(red)(9^(1/6))=color(red)((3^2)^(1/6))=color(red)(3^(2xx1/6))=color(red)(3^(2/6))=color(red)(3^(1/3))=color(red)(root3(3)#

#(-root3(3))/((4sqrt(2)sqrt2)xx(root3(3)))#

#(-root3(3))/(4(2)xxroot3(3)#

And here's what we were waiting for (it actually set itself up last step but I wanted to clean things up first before dealing with the cancellation here)

#(-cancelroot3(3))/(8xxcancelroot3(3)#

#-1/8#