irrational fraction

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spacewater

Junior Member
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Jul 10, 2009
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problem : 16 over 16 cube root
\(\displaystyle \frac{16}{^3\sqrt{16}}\)
steps
\(\displaystyle \frac{16 \cdot ^3\sqrt{16}}{^3\sqrt{16} \cdot ^3\sqrt{16}}\)


\(\displaystyle \frac{16 \cdot ^3\sqrt{16}}{16}\)

Cancel out 16 on the numerator and the denominator

the final answer

\(\displaystyle 2\cdot^3\sqrt{2}\)
my final answer does not correspond with the answersheet. I believe i did everything right so i am seeking for help.

problem 2: 6 over 1 minus 3 square root
\(\displaystyle \frac{6}{1 - \sqrt{3}}\)

steps
\(\displaystyle \frac {6\cdot 1+\sqrt{3}}{1-\sqrt{3}\cdot 1+\sqrt{3}}\)

\(\displaystyle \frac{6+6\sqrt{3}}{1-3}\)

final answer
\(\displaystyle -\frac{6-6\sqrt{3}}{2}\)

Shouldn't this be the final answer since you cant cancel out the numerator and denominator due to the subtraction between \(\displaystyle 6\) and \(\displaystyle 6\sqrt{3}\)? I thought only the one with either multiplication/division can cancel out when it comes to fraction.
For example
\(\displaystyle \frac{6+x}{2x}\)
2 and 6 shouldn't cancel out since there is an addition going on in the numerator.
 
#1. \(\displaystyle \frac{16}{16^{\frac{1}{3}}}=16^{\frac{2}{3}}=\sqrt[3]{16^{2}}=\sqrt[3]{256}=\sqrt[3]{64\cdot 4}=4\cdot 4^{\frac{1}{3}}=4\cdot 2^{\frac{2}{3}}\)

There is one way to look at it.


Or multiply top and bottom by \(\displaystyle 16^{\frac{2}{3}}\) to get rid of the radical in the denominator.

\(\displaystyle \frac{16}{16^{\frac{1}{3}}}\cdot\frac{16^{\frac{2}{3}}}{16^{\frac{2}{3}}}=\frac{16^{\frac{5}{3}}}{16}=16^{\frac{2}{3}}=4\cdot 2^{\frac{2}{3}}\)
 
spacewater said:
problem : 16 over 16 cube root
\(\displaystyle \frac{16}{^3\sqrt{16}}\)
steps
\(\displaystyle \frac{16 \cdot ^3\sqrt{16}}{^3\sqrt{16} \cdot ^3\sqrt{16}}\)

\(\displaystyle \frac{16 \cdot ^3\sqrt{16}}{16}\)
___________________________________________________________________
Step above is incorrect because:

\(\displaystyle ^3\sqrt{16} \cdot ^3\sqrt{16} \, = \, ^3\sqrt{16^2} \, = 16^{\frac{2}{3}\)
______________________________________________________
Cancel out 16 on the numerator and the denominator

the final answer

\(\displaystyle 2\cdot^3\sqrt{2}\)
my final answer does not correspond with the answersheet. I believe i did everything right so i am seeking for help.
____________________________________________________________________
If I were to do this problem:\(\displaystyle \frac{16}{^3\sqrt{16}}\)

\(\displaystyle = \frac{2^4}{^3\sqrt{2^4}}\)

\(\displaystyle = \frac{2^4}{2^{\frac{4}{3}}}\)

\(\displaystyle = \, 2^{(4 \, - \, \frac{4}{3})} \, = \, 2^{2\frac{2}{3}} = \, 4 \cdot ^3\sqrt{4}\)
_______________________________________________________________
problem 2: 6 over 1 minus 3 square root
\(\displaystyle \frac{6}{1 - \sqrt{3}}\)

steps
\(\displaystyle \frac {6\cdot 1+\sqrt{3}}{1-\sqrt{3}\cdot 1+\sqrt{3})}\)

\(\displaystyle \frac{6+6\sqrt{3}}{1-3}\)

final answer
\(\displaystyle -\frac{6-6\sqrt{3}}{2}\)

Shouldn't this be the final answer since you cant cancel out the numerator and denominator due to the subtraction between \(\displaystyle 6\) and \(\displaystyle 6\sqrt{3}\)? I thought only the one with either multiplication/division can cancel out when it comes to fraction.
For example
\(\displaystyle \frac{6+x}{2x}\)
2 and 6 shouldn't cancel out since there is an addition going on in the numerator.
Click to expand...
 
\(\displaystyle \sqrt[3]{16}\cdot\sqrt[3]{16}=\sqrt[3]{256}=4\sqrt[3]{4}\cdot\)

\(\displaystyle \sqrt[3]{16}\cdot\sqrt[3]{16}\cdot\sqrt[3]{16}=16\)

Try this...
\(\displaystyle \frac{16}{\sqrt[3]{16}}=\frac{16}{2\sqrt[3]{2}}\cdot\frac{\sqrt[3]{4}}{\sqrt[3]{4}}=\)???
 
The reason your answer doesn't correspond to the correct one is that you have said \(\displaystyle \sqrt[3]{16}\cdot\sqrt[3]{16}=16\) which is incorrect. These are cube roots, not square roots.
 
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