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anonymous

  • one year ago

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

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  1. kittiwitti1
    • one year ago
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    "Join"?

  2. anonymous
    • one year ago
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    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
    • one year ago
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    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
    • one year ago
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    yes

  5. kittiwitti1
    • one year ago
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    Okay... do you know how to simplify the exponent 1/3 of the bottom portion?

  6. anonymous
    • one year ago
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    i tried multiplying by the bottom conjugate but i got very lost.

  7. anonymous
    • one year ago
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    by cubing it

  8. kittiwitti1
    • one year ago
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    https://www.mathsisfun.com/algebra/exponent-fractional.html Have you tried this?

  9. anonymous
    • one year ago
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    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
    • one year ago
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    Well let's convert the x^(1/3) first.

  11. kittiwitti1
    • one year ago
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    \[x^{\frac{a}{b}}=\sqrt[b]{x^{a}}\]\[x^{\frac{1}{3}}=\sqrt[3]{x^{1}}\]

  12. anonymous
    • one year ago
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    ok i get that part

  13. kittiwitti1
    • one year ago
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    Okay so now we have\[\frac{(x-27)}{\sqrt[3]{x}-3}\]

  14. kittiwitti1
    • one year ago
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    Hmmm, does this help? http://openstudy.com/study#/updates/4e735b990b8b247045d029a1

  15. anonymous
    • one year ago
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    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
    • one year ago
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    Try the link I posted.

  17. kittiwitti1
    • one year ago
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    Does it help?

  18. anonymous
    • one year ago
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    yes but i didn't know even without the cubed root with x it qualified as the sum of cube formula

  19. kittiwitti1
    • one year ago
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    http://prntscr.com/8g4uw1 Mathematical terminology of "join": to connect with a straight line or curve. Does this help?

  20. anonymous
    • one year ago
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    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
    • one year ago
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    @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
    • one year ago
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    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
    • one year ago
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    Apparently they found a way to use exponent 1/3 to make the process easier.

  24. kittiwitti1
    • one year ago
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    \[\frac{x-27}{\sqrt[3]{x}-3}\rightarrow\frac{x^{1/3}-3{x^{1/3}+3x^{1/3}+9}}{\sqrt[3]{x}-3}\]...I think I messed up somewhere.

  25. anonymous
    • one year ago
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    you dont distribute it, you would cancel then \[x ^{1/3}-3\] from the top and bottom

  26. anonymous
    • one year ago
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    so it would be \[((x ^{1/3}-3)(x ^{2/3}+(3timesx ^{1/3})+9)div(x ^{1/3}-3)\]

  27. kittiwitti1
    • one year ago
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    Sorry my mom was cussing me out for being on OS helping people with math ^^;

  28. kittiwitti1
    • one year ago
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    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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