## anonymous one year ago Join (x-27)/(x^(1/3)-3)

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1. kittiwitti1

"Join"?

2. 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

3. 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?

4. anonymous

yes

5. kittiwitti1

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

6. anonymous

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

7. anonymous

by cubing it

8. kittiwitti1

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

9. anonymous

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

10. kittiwitti1

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

11. kittiwitti1

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

12. anonymous

ok i get that part

13. kittiwitti1

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

14. kittiwitti1

15. 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

16. kittiwitti1

17. kittiwitti1

Does it help?

18. anonymous

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

19. kittiwitti1

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

20. 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 $(x ^{1/3}-3)(x ^{2/3}+x ^{1/3}3+9$

21. 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)...?

22. 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

23. kittiwitti1

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

24. 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.

25. anonymous

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

26. anonymous

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

27. kittiwitti1

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

28. 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!