Rationalize the denominator: cubic root of 2y^4/6x^4

Question

anonymous · Answer

$$\sqrt[3]{2y ^{4}/6x ^{4}}$$

anonymous · Answer

My instinct is to go down instead of making it $$x ^{64}$$ because that would look dumb, unless it's correct but I've never had to decrease... do I just multiply by a negative exponent?

anonymous · Answer

=cube root (2y^4)/cube root(6x^4)
=cube root(2y^4)*cuberoot(6x^4)/cube root(6x^4)*cube root(6x^4)
=cube root(2y^4)*cuberoot(6x^4)/6x^4

basically separate top and bottom roots
multiply by conjugates
fix ugly maths

jhonyy9 · Answer

cuberoot(y/3x)^4 =(y/3x)cuberoot(y/3x)=(ycuberoot3xy)/3x

jhonyy9 · Answer

2/6 simplifie by 2 and get 1/3

jhonyy9 · Answer

ok ?

anonymous · Answer

Hmm, well that is handy, didn't know oyu could simplify it but what would the correct conjugate be?

anonymous · Answer

you*

jhonyy9 · Answer

good luck 

bye

jhonyy9 · Answer

1/sqrt3 =(sqrt3)/3 ---like example

jhonyy9 · Answer

ok ?

anonymous · Answer

No actually, I'm confused... can you just tell me what I need to multiply by to get rid of the radical in the denominator?

jhonyy9 · Answer

1/sqrt3 so when there is radical in denominator multiplie numerator and denominator by conjugate so in this case by sqrt3 and in this way you get in denominator 3 and in numerator sqrt3

jhonyy9 · Answer

because sqrt3 * sqrt3 =3

anonymous · Answer

I know... but what would the conjugate be. I know how to do all this already i get it I just need to know if the conjugate I am using is correct...

jhonyy9 · Answer

your answer not is right

jhonyy9 · Answer

check my answer step by step

razor99 · Answer

we have to simplify that.

anonymous · Answer

Ok look, the denominator is cubic root of 6x^4 .... so what do I multiply it by to take a perfect cubic root?

anonymous · Answer

sorry that is wrong

anonymous · Answer

$$\sqrt[3]{36x^2}$$

saifoo.khan · Answer

If sat is here, then there was no need of calling me! :)

anonymous · Answer

then you will get in the denominator a perfect cube, namely 
$$\sqrt[3]{216x^6}$$

anonymous · Answer

He just got here thanks anyway

anonymous · Answer

you got this?

anonymous · Answer

you have 
$$\sqrt[3]{6x^4}$$ so to make it a perfect cube, you need the exponent on the x to be 6, and you have to multiply by 6^2=36  to make the constnant a cube

anonymous · Answer

Wait why does the exponent for x have to be 6?

anonymous · Answer

because you have a cube root so you need a number divisible by 3

anonymous · Answer

i mean an exponent divisible by 3

anonymous · Answer

$$\sqrt[3]{216x^6}=6x^2$$

anonymous · Answer

So that would come out as $$6x ^{2}$$ ?

anonymous · Answer

yes

anonymous · Answer

Ah thank you that is ALL I needed, I guess nobody understood what I meant. Thank you very much

anonymous · Answer

yw