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

the final answer

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}

final answer

-\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.

\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

the final answer

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}

final answer

-\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.