- anonymous

Join (x-27)/(x^(1/3)-3)

- katieb

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- kittiwitti1

"Join"?

- anonymous

I don't know what that means, I was hoping someone would. But I googled it and all it took me to was to joining fraction. A website gave me this answer\[x ^{2/3}+3\sqrt[3]{x}+9\] but I don't know how they got there

- kittiwitti1

hmm I'll type it out in a more readable format first then\[\frac{\left(x-27\right)}{\left(x^{\frac{1}{3}}-3\right)}\]is this right?

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## More answers

- anonymous

yes

- kittiwitti1

Okay... do you know how to simplify the exponent 1/3 of the bottom portion?

- anonymous

i tried multiplying by the bottom conjugate but i got very lost.

- anonymous

by cubing it

- kittiwitti1

https://www.mathsisfun.com/algebra/exponent-fractional.html
Have you tried this?

- anonymous

i went on the page and i get the information, I just don't get how to tie it in with the problem

- kittiwitti1

Well let's convert the x^(1/3) first.

- kittiwitti1

\[x^{\frac{a}{b}}=\sqrt[b]{x^{a}}\]\[x^{\frac{1}{3}}=\sqrt[3]{x^{1}}\]

- anonymous

ok i get that part

- kittiwitti1

Okay so now we have\[\frac{(x-27)}{\sqrt[3]{x}-3}\]

- kittiwitti1

Hmmm, does this help?
http://openstudy.com/study#/updates/4e735b990b8b247045d029a1

- anonymous

the problem is \[(x-27)\div(\sqrt[3]{x}-3)\] but it simplifies to \[\sqrt[3]{x}^{2}+3\sqrt[3]{x}+9\], i dont understand how to simplify it

- kittiwitti1

Try the link I posted.

- kittiwitti1

Does it help?

- anonymous

yes but i didn't know even without the cubed root with x it qualified as the sum of cube formula

- kittiwitti1

http://prntscr.com/8g4uw1
Mathematical terminology of "join": to connect with a straight line or curve.
Does this help?

- anonymous

the regular cube formula is\[(x-3)(x ^{2}+3x+9)\]right what i didn't understand is that you can cube root it so it would turn out to \[(x ^{1/3}-3)(x ^{2/3}+x ^{1/3}3+9\]

- kittiwitti1

@ganeshie8
(Sorry, I'm a tad muddled in the brain here. I may need correcting)
From what I can understand they simply substituted the exponent for an x^(1/3)...?

- anonymous

yes because \[x ^{3}-27\] is a perfect cube and i'm assuming that x-27 is what it look like when its cubed rooted

- kittiwitti1

Apparently they found a way to use exponent 1/3 to make the process easier.

- kittiwitti1

\[\frac{x-27}{\sqrt[3]{x}-3}\rightarrow\frac{x^{1/3}-3{x^{1/3}+3x^{1/3}+9}}{\sqrt[3]{x}-3}\]...I think I messed up somewhere.

- anonymous

you dont distribute it, you would cancel then \[x ^{1/3}-3\] from the top and bottom

- anonymous

so it would be \[((x ^{1/3}-3)(x ^{2/3}+(3timesx ^{1/3})+9)div(x ^{1/3}-3)\]

- kittiwitti1

Sorry my mom was cussing me out for being on OS helping people with math ^^;

- kittiwitti1

It may be worth noting that \[\sqrt[3]{27}=3\]Reason:\[27=3\times3\times3\]I'm stuck here, sorry. Maybe ask @ganeshie8 for help e_e!

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