Question

What is 3 root of 384?

A.
3 3 root of 6

B.
4 3 root of 3

C.
4 3 root of 6

D.
6 3 root of 3

Answer 1

it is supposed to look like this \[a.3\sqrt[3]{6}\]

Answer 2

\[b.4\sqrt[3]{3}\]

Answer 3

\[c.4\sqrt[3]{6}\]

Answer 4

\[d.6\sqrt[3]{3}\]

Answer 5

you can test each answer, by "moving" the number on the outside of the cube root to inside *but* put in 3 copies
for example
\[ 3\sqrt[3]{6} = \sqrt[3]{6\cdot 3\cdot 3\cdot 3 } \]
is that the original problem? if we multiply it out we get
\[ 3\sqrt[3]{6} = \sqrt[3]{6\cdot 3\cdot 3\cdot 3 } = \sqrt[3]{162}\]
which is not the original problem. so the answer is not A

Answer 6

can you do that for the other choices?

Answer 7

how would i do c like this or \[\sqrt[3]{6*4*4*4*4}\]

Answer 8

or is that not right

Answer 9

4 times itself 3 times (not 4 times)
we are using the idea that
\[4= \sqrt[3]{4\cdot 4 \cdot 4} \]
to "move the 4 inside"

Answer 10

do i take the 4th 4 away because then it = 384

Answer 11

wait so is it c then

Answer 12

The 4th 4 does not belong there.
\[ 4\sqrt[3]{6} = \sqrt[3]{6 \cdot 4 \cdot 4 \cdot 4} \]

Answer 13

okay so it is c

Answer 14

yes

Answer 15

the other way to do the problem is factor 384 into its prime factors
2*192= 2*2*96= 2*2*2*48= 2*2*2*2*24 = 2*2*2*2*2*12= 2*2*2*2*2*2*2*3
and pull out triples (2*2*2) * (2*2*2) * 2*3
in other words
\[ \sqrt[3]{ 2 \cdot 2 \cdot 2} \sqrt[3]{ 2 \cdot 2 \cdot 2} \sqrt[3]{ 2 \cdot 3} \]
which becomes
\[ 2 \cdot 2 \sqrt[3]{ 6} \\ 4 \sqrt[3]{ 6}\]

Answer 16

okay thanks