Question

Help me solve (^6 sqrt of x)^3?

Answer 1

It originally equals x^1/3

Answer 2

But the question is: \[(\sqrt[6]{x})^3\]

Answer 3

It's the parenthesis that throws me off..

Answer 4

You can, of course, express everything as exponents first. That makes easier (to figure out)

In particular, sixth root simply means an exponent of 1/6.
\[\LARGE \sqrt[6]x= \color{blue}{x^{\frac16}}\]

In effect,
\[\LARGE (\sqrt[6]x)^3 = \left(\color{blue}{x^{\frac16}}\right)^3\]
Which may now be simplified according to the laws of exponents...

Answer 5

I already wrote out the rule to you @whalexnuker

Answer 6

It's the same format

Answer 7

Perhaps that might help me better... I didn't think of simplifying it from that. Lol <_> I'm still getting a result of sqrt of x.
the parenthesis are still throwing me off. Solving it on my own (because i was using a program), i would like to say you'd multiply everything on the inside with the 3?
Yeah, hero. Let me look at that because the problem is the same LOL. i just noticed. But since I have no C, it's just..... x^3/6?

Answer 8

Wait, sorry. I don't have A. but the equation is the same

Answer 9

it just the parenthesis that still throws me off... I realize its the same as the other but I don't have A so nothing is being multiplied..

Answer 10

is it x^1/2? am I dividing the 3 with the 6?

Answer 11

riiiggghhhtt? >.> @Hero

Answer 12

correct, but @terenzreignz helped you with this. Give the medal to him.

Answer 13

Done :D I couldn't have done it without that simplified form. I would have been stuck on this forever. Thanks.

Answer 14

What about ^5 sqrt (x) * (x^2/5) * (x^3/5) * (x^4/5) @terenzreignz >.> since the denominators are the same outside of the square root, i can add the exponents together, yes? 2+3+5 = x^10/5
^5sqrt(x) x^10/5

Answer 15

With the 5 that's on the input, does that get divided by the 10/5 or something?

Answer 16

it equals x^2 but I'm trying to figure out how.

Answer 17

@bhaskar.pathk Are you just going to sit on my question....? you've been here for a bit...

Answer 18

Sorry, was spaced out... hang on...

Answer 19

But didn't you already figure it out?
Adding the exponents gives 1/5 + 2/5 + 3/5 + 4/5 = 10/5 = 2

Answer 20

Not forgetting, of course, that \[\LARGE \sqrt[5]x = x^{\frac15 }\]

Answer 21

@whalexnuker : sorry was facing some network prblm

Answer 22

I think I solved half of it :p I got the 10/5 which equals 2 but I still have the ^5 sqrt of (x).
So, right now, it's ^5sqrt of (x) 10/5 (or 2).
I dunno what to do with that "^5" since the square root is only covering the X and not the 2

Answer 23

like this:
\[\Large \sqrt[5]x \cdot x^{\frac25}\cdot x^{\frac35}\cdot x^{\frac45}\]

Answer 24

But because of this fact:
\[\LARGE \sqrt[5]x = x^{\frac15 }\]

That expression on top is just equal to \[\Large x^{\frac15} \cdot x^{\frac25}\cdot x^{\frac35}\cdot x^{\frac45}\]
And now, the exponents may be easily added to get the desired result.

Answer 25

like this:
\[\Large \sqrt[5]x \cdot x^{\frac25}\cdot x^{\frac35}\cdot x^{\frac45}\]

Answer 26

!! oh! See, I forgot to switch it again XD I can now just add them! Awesome! and THAT gives me x^10/5 = x^2
I'd give you medals if I could. In fact, let me scroll through your profile and I will :D

Answer 27

No, there's no need for that, as long as you got it, that's enough :)

Answer 28

It's just duh moments >.> My whole lesson is about converting them like that and I let it completely slip my mind lmao. Thanks so much.

Answer 29

When in doubt, turn everything into exponents, fractional or not, they're generally easier to work with.

Answer 30

...than that nasty radical sign \(\Large \sqrt{\qquad}\)

Answer 31

I need to sign off now... good luck with the rest (if any)
^_^
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Terence out