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

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

Mathematics
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help plz

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  • phi
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 \[ \frac{\sqrt[3]{2} \sqrt[3]{6} \sqrt[3]{6} }{\sqrt[3]{6} \sqrt[3]{6} \sqrt[3]{6} } \]
  • phi
by definition, the bottom becomes 6 (the cube root multiplied three times undoes the radical)
Why would you multiply the numerator and the denominator twice by the denominator
oh ok
because it is cubed
  • phi
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 \[ \frac{2 \sqrt[3]{3\cdot 3}}{6} \]
nvmd
and then multiply the 3's on the inside of the radical
  • phi
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
so it would be A.
  • phi
yes
ok I understand how to do it now thankyou very much
  • phi
yw

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