Question

last question...

Answer 1

question:\[\sqrt[3]{54}+\sqrt[3]{16}-\sqrt[3]{24}\]

Answer 2

factor each number under each radical and take out all cube powers

Answer 3

i.e. what is the prime factorization of 16 ?

Answer 4

4?

Answer 5

But they are not perfect cubes.

Answer 6

2*2*2*2

Answer 7

and that is not the prime factorization
do you know how to factor a number?

Answer 8

my last comment was at @bridgetolivares

Answer 9

do you see what pratu did there bridget?

Answer 10

nope

Answer 11

if you don't know how to factor a number you have been skipping out on math for some time, and that makes these problems very difficult to solve
you need to know how to get the \(prime\) \(factorization\) of the number under the sign
http://www.mathsisfun.com/prime-factorization.html

Answer 12

so what we are doing is breaking the number down into multiples of prime numbers
16 is even, so it is clearly divisible by 2
so that is the first thing we do: write
16=2*8

Answer 13

ooooh i understand what youre saying now.

Answer 14

but 8 is not prime, and because it is even we can again divide it by 2 and rewrite 16 as
16=2*8=2*2*4
again doing this to the 4 we get the final prime factorization:
16=2*2*2*2
now what does this mean for us?

Answer 15

we want the cube root of this (3rd root)
so for every number we have three copies of, we take out one

Answer 16

16=2*2*2*2=2*(2*2*2)=2*(2^3)
so this is how it works now:\[\sqrt[3]{16}=\sqrt[3]{2\cdot2\cdot2\cdot2}=\sqrt[3]{(2\cdot2\cdot2)\cdot2}=\sqrt[3]{2^3\cdot2}=2\sqrt[3]2\]please do make sure you understand each part of that, not that you need to write it out that way every time

Answer 17

wait wouldnt the answer be \[4\sqrt[3]{2}\]

Answer 18

why should it be a 4 and not a 2 ?

we can only take out three copies of a number, and put that number out front.

since we have four 2's, we can take out three of them, and we have to leave the fourth one under the radical

Answer 19

\[\sqrt[3]{16}=\sqrt[3]{2\cdot2\cdot2\cdot2}=\sqrt[3]{(2\cdot2\cdot2)\cdot2}=\sqrt[3]{2^3\cdot2}=2\sqrt[3]2\] ^^^^ ^
this is where the 2 comes from

Answer 20

oh ok. but that answer isnt one of the options

Answer 21

I was only talking about the middle term, \(\sqrt[3]{16}\)
your job is to do the other two :)

Answer 22

wait nvm. yeah i had forgotten.

Answer 23

so why don't you go ahead and try to get the prime factorization of the number in the third term 24
?

Answer 24

iim working on it now

Answer 25

good luck :)