Question

I need help with this problem trying to figure out how to do it, plz help. I will give medal.

Answer 1

help plz

Answer 2

@phi

Answer 3

if you have a "cube root" then if you multiply it by itself *three times* , you "get rid" of the the radical

when they say "rationalized denominator" it means get rid of the radical in the bottom part of the fraction

so to do that, multiply the bottom by \(\sqrt[3]{6} \sqrt[3]{6} \)
and to keep things equal , you have to multiply the top by the same thing
you get
\[ \frac{\sqrt[3]{2} \sqrt[3]{6} \sqrt[3]{6} }{\sqrt[3]{6} \sqrt[3]{6} \sqrt[3]{6} } \]

Answer 4

by definition, the bottom becomes 6
(the cube root multiplied three times undoes the radical)

Answer 5

Why would you multiply the numerator and the denominator twice by the denominator

Answer 6

oh ok

Answer 7

because it is cubed

Answer 8

in the top, all three are cube roots so you can combine the top into
\[ \sqrt[3]{2\cdot 6 \cdot 6} \]
I would factor the 6 into 2*3 so we can write the top as
\[ \sqrt[3]{2\cdot 2 \cdot 3 \cdot 2 \cdot 3} \]
or reordering to make it clearer
\[ \sqrt[3]{2\cdot 2 \cdot 2 \cdot 3 \cdot 3} \]

now if we find *three of the same term* inside a cube root. we can "pull them out"
and replace them with 1 term on the outside
\[2 \sqrt[3]{3\cdot 3} \]

so far we have
\[ \frac{2 \sqrt[3]{3\cdot 3}}{6} \]

Answer 9

nvmd

Answer 10

and then multiply the 3's on the inside of the radical

Answer 11

can you finish?
we can't do much with the 3's inside (we would need 3 of them to simplify)
so I would make them 9
and of course we can simplify 2/6 on the outside

Answer 12

so it would be A.

Answer 13

yes

Answer 14

ok I understand how to do it now thankyou very much

Answer 15

yw