Question

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

Answer 1

"Join"?

Answer 2

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

Answer 3

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?

Answer 4

yes

Answer 5

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

Answer 6

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

Answer 7

by cubing it

Answer 8

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

Answer 9

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

Answer 10

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

Answer 11

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

Answer 12

ok i get that part

Answer 13

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

Answer 14

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

Answer 15

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

Answer 16

Try the link I posted.

Answer 17

Does it help?

Answer 18

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

Answer 19

http://prntscr.com/8g4uw1

Mathematical terminology of "join": to connect with a straight line or curve.
Does this help?

Answer 20

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 \[(x ^{1/3}-3)(x ^{2/3}+x ^{1/3}3+9\]

Answer 21

@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)...?

Answer 22

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

Answer 23

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

Answer 24

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

Answer 25

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

Answer 26

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

Answer 27

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

Answer 28

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!