Question

I'm needing some help with this "Cube root" problem. Please show me how to do this, I've been stuck erasing this off my dry erase board over and over and haven't been able to manage it

The cube root of 54x to the 8th power.

Answer 1

I'm sorry, I should clarify. They're asking me to simplify
"The cube root of 54x to the 8th power"

Answer 2

\[\sqrt[3]{54x^8}\]

Answer 3

like that?

Answer 4

Yes, exactly. Sorry, I don't exactly know how to write it like that in this text box.

Answer 5

But they're asking me to simplify that problem.

Answer 6

And here's where I get stuck.
I turn "The cube root of 54x^8"
into "the cube root of 27x^7 times 2x"
Then I split that apart. So it becomes the cube root of 27x, times the cube root of 2x. And I'm not sure where to go from there. Or "Why"

Answer 7

54=9*6=3*3*3*2 since its a cube root we need 3 of a kind so we can take a 3 and write it on the outside as for the x^8, 8/3=2 remainder 2, so x^2 can come outside the root but an x^2 must stay under, final answer : \[3x^{2}\sqrt[3]{2x^{2}}\]

Answer 8

do number by themselves then letters with exponents

Answer 9

How does "8/3=2 remainder 2" happen?

Answer 10

Oh wait, that's silly. I'm sorry

Answer 11

I'm putting a bit too much thought into it then I should I guess. Thanks for the help! I think it's starting to click with me now.

Answer 12

when you have a letter such as x^8 to see how many x's can be written on the outside you must divide by whatever root it is (square, cube, 4th root etc) so 8/3 =2 so X^2 can be written on the outside however the remainder of 8/3=2 so x^2 stays under the root as well

Answer 13

example \[\sqrt[4]{y^{16}h^{5}g^{7}}\]

Answer 14

16/4=4 so y^4 written on outside of root, 5/4= 1 and remainder 1 therefore h^1 written on outside and h^1 remains under root, 7/4=1 remainder 3 so g^3 stays under root leaving us with :\[y^{4}h^{1}g^{1}\sqrt[4]{h^{1}g^{3}}\]