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spacewater
07-16-2009, 10:58 AM
problem : 16 over 16 cube root
\frac{16}{^3\sqrt{16}}
steps
\frac{16 \cdot ^3\sqrt{16}}{^3\sqrt{16} \cdot ^3\sqrt{16}}

\frac{16 \cdot ^3\sqrt{16}}{16}

Cancel out 16 on the numerator and the denominator

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
\frac{6}{1 - \sqrt{3}}

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

\frac{6+6\sqrt{3}}{1-3}

-\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 6 and 6\sqrt{3}? I thought only the one with either multiplication/division can cancel out when it comes to fraction.
For example
\frac{6+x}{2x}
2 and 6 shouldn't cancel out since there is an addition going on in the numerator.

galactus
07-16-2009, 11:06 AM
#1. \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 16^{\frac{2}{3}} to get rid of the radical in the denominator.

\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}}

Subhotosh Khan
07-16-2009, 11:51 AM
problem : 16 over 16 cube root
\frac{16}{^3\sqrt{16}}
steps
\frac{16 \cdot ^3\sqrt{16}}{^3\sqrt{16} \cdot ^3\sqrt{16}}

\frac{16 \cdot ^3\sqrt{16}}{16}
__________________________________________________ _________________
Step above is incorrect because:

^3\sqrt{16} \cdot ^3\sqrt{16} \, = \, ^3\sqrt{16^2} \, = 16^{\frac{2}{3}
__________________________________________________ ____
Cancel out 16 on the numerator and the denominator

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:\frac{16}{^3\sqrt{16}}

= \frac{2^4}{^3\sqrt{2^4}}

= \frac{2^4}{2^{\frac{4}{3}}}

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

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

\frac{6+6\sqrt{3}}{1-3}

-\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 6 and 6\sqrt{3}? I thought only the one with either multiplication/division can cancel out when it comes to fraction.
For example
\frac{6+x}{2x}
2 and 6 shouldn't cancel out since there is an addition going on in the numerator.

Loren
07-16-2009, 11:54 AM
\sqrt[3]{16}\cdot\sqrt[3]{16}=\sqrt[3]{256}=4\sqrt[3]{4}\cdot

\sqrt[3]{16}\cdot\sqrt[3]{16}\cdot\sqrt[3]{16}=16

Try this...
\frac{16}{\sqrt[3]{16}}=\frac{16}{2\sqrt[3]{2}}\cdot\frac{\sqrt[3]{4}}{\sqrt[3]{4}}=???

DrMike
07-20-2009, 08:39 PM
The reason your answer doesn't correspond to the correct one is that you have said \sqrt[3]{16}\cdot\sqrt[3]{16}=16 which is incorrect. These are cube roots, not square roots.