SOLUTION: why is the answer to cube root of 9 times cubed root of 6 DIVIDED BY 6 root 2 times 6 root 2 = 3?

Algebra ->  Square-cubic-other-roots -> SOLUTION: why is the answer to cube root of 9 times cubed root of 6 DIVIDED BY 6 root 2 times 6 root 2 = 3?      Log On


   

Question 395152: why is the answer to
cube root of 9 times cubed root of 6
DIVIDED BY
6 root 2 times 6 root 2
= 3?
Answer by jsmallt9(3758) About Me  (Show Source):
You can put this solution on YOUR website!
%28root%283%2C+9%29%2Aroot%283%2C+6%29%29%2F%28root%286%2C+2%29%2Aroot%286%2C+2%29%29
First we can use a property of radicals, root%28a%2C+p%29%2Aroot%28a%2C+q%29+=+root%28a%2C+p%2Aq%29, to multiply in both the numerator and the denominator:
root%283%2C+9%2A6%29%2Froot%286%2C+2%2A2%29
or
root%283%2C+54%29%2Froot%286%2C+2%5E2%29
I left the radicand in the denominator in the form of 2%5E2 for reasons that will become clear soon.
Next we use fractional exponents in the denominator to help us simplify further. Using the pattern root%28a%2C+p%5Eb%29+=+a%5E%28b%2Fa%29 we can rewrite the denominator as:
2%5E%282%2F6%29
We can see that this fraction will reduce to:
2%5E%281%2F3%29
Writing this back in radical form we get:
root%283%2C+2%5E1%29
or
root%283%2C+2%29
Now our fraction looks like:
root%283%2C+54%29%2Froot%283%2C+2%29
Now we can use another property of radicals, root%28a%2C+p%29%2Froot%28a%2C+q%29++=+root%28a%2C+p%2Fq%29 to combine the two radicals into one:
root%283%2C+54%2F2%29
which simplifies to:
root%283%2C+27%29
And since 3%5E3+=+27, this cube root becomes a 3!