anonymous
  • anonymous
Join (x-27)/(x^(1/3)-3)
Calculus1
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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kittiwitti1
  • kittiwitti1
"Join"?
anonymous
  • 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
  • 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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anonymous
  • anonymous
yes
kittiwitti1
  • kittiwitti1
Okay... do you know how to simplify the exponent 1/3 of the bottom portion?
anonymous
  • anonymous
i tried multiplying by the bottom conjugate but i got very lost.
anonymous
  • anonymous
by cubing it
kittiwitti1
  • kittiwitti1
https://www.mathsisfun.com/algebra/exponent-fractional.html Have you tried this?
anonymous
  • 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
  • kittiwitti1
Well let's convert the x^(1/3) first.
kittiwitti1
  • kittiwitti1
\[x^{\frac{a}{b}}=\sqrt[b]{x^{a}}\]\[x^{\frac{1}{3}}=\sqrt[3]{x^{1}}\]
anonymous
  • anonymous
ok i get that part
kittiwitti1
  • kittiwitti1
Okay so now we have\[\frac{(x-27)}{\sqrt[3]{x}-3}\]
kittiwitti1
  • kittiwitti1
anonymous
  • 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
  • kittiwitti1
Try the link I posted.
kittiwitti1
  • kittiwitti1
Does it help?
anonymous
  • anonymous
yes but i didn't know even without the cubed root with x it qualified as the sum of cube formula
kittiwitti1
  • kittiwitti1
http://prntscr.com/8g4uw1 Mathematical terminology of "join": to connect with a straight line or curve. Does this help?
anonymous
  • 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
  • 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
  • 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
  • kittiwitti1
Apparently they found a way to use exponent 1/3 to make the process easier.
kittiwitti1
  • 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
  • anonymous
you dont distribute it, you would cancel then \[x ^{1/3}-3\] from the top and bottom
anonymous
  • anonymous
so it would be \[((x ^{1/3}-3)(x ^{2/3}+(3timesx ^{1/3})+9)div(x ^{1/3}-3)\]
kittiwitti1
  • kittiwitti1
Sorry my mom was cussing me out for being on OS helping people with math ^^;
kittiwitti1
  • 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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