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TeemoTheTerific

hi Can someone help me with a question? Help much appreciated

  • one year ago
  • one year ago

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  1. Razzputin
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    hello

    • one year ago
  2. TeemoTheTerific
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    \[ \huge x=2^{\frac{ 1 }{ 3 }} + 4^{\frac{ 1 }{ 3 }}\] Show that \[\huge x ^{3} = 6(1+x)\]

    • one year ago
  3. TeemoTheTerific
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    can you please show me all woking out as i tried it and it took up like half a page :P(prob wrong working out)

    • one year ago
  4. Razzputin
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    post your working out, ill look at it and spot the error.

    • one year ago
  5. TeemoTheTerific
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    i worked out the Right hand side is equal to 46

    • one year ago
  6. Razzputin
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    have you got a scanner?

    • one year ago
  7. TeemoTheTerific
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    no...

    • one year ago
  8. Razzputin
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    damn, have you got any way of reproducing the page of working you did so I can look at it?

    • one year ago
  9. TeemoTheTerific
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    yea ok... OH YEA the right hand side isnt equal to 46

    • one year ago
  10. TeemoTheTerific
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    |dw:1359973881754:dw|

    • one year ago
  11. TeemoTheTerific
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    now i dont think you can do anything with the RHS

    • one year ago
  12. TeemoTheTerific
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    |dw:1359973981397:dw|

    • one year ago
  13. TeemoTheTerific
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    im pretty sure there is a shortcut to do this question?

    • one year ago
  14. Razzputin
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    do you get x^3 = 23.08393261?

    • one year ago
  15. TeemoTheTerific
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    im pretty sure we need to show it using indices

    • one year ago
  16. Razzputin
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    lol, i dont even know those. As an answer i get \[\pm 2.847322102\]

    • one year ago
  17. TeemoTheTerific
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    well i now completely lost with this quesiton :P

    • one year ago
  18. Razzputin
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    you were going the right way.\[x^{3} = 6(1+(2^\frac{ 1 }{ 3 }+4^\frac{ 1 }{ 3 }))\] \[x^{3} = 6(1+(\sqrt[3]{2}+\sqrt[3]{4}))\]\[x^{3} = 6(1+(2.847322102))\]\[x^{3} = 23.08393261\]\[x =\sqrt[3]{23.08393261}\]\[x = \pm2.84732210\]

    • one year ago
  19. TeemoTheTerific
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    we are meant to prove that \[x ^{3} = 6(1+x)\] not find x

    • one year ago
  20. Jonask
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    \[x^3+y^3=(x+y)(x^2-xy+y^2)\] \[\huge x^3=(2+4)(2^{\frac { 2 }{ 3 }}-8^{\frac{ 1 }{ 3 }}+4^{\frac{ 2 }{ 3 }})\]

    • one year ago
  21. Razzputin
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    oh blast. lol well then if x^3 = 23.08393261 then substitute given in for x therefore \[(\sqrt[3]{2}+\sqrt[3]{4})^{3}\] what does it equal?

    • one year ago
  22. Jonask
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    we not meant to find x but prove

    • one year ago
  23. Razzputin
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    its basic substitution

    • one year ago
  24. TeemoTheTerific
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    \[\huge =6(2^{\frac { 2 }{ 3 }}-2+4^{\frac{ 2 }{ 3 }})\]

    • one year ago
  25. Jonask
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    yes

    • one year ago
  26. TeemoTheTerific
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    yea but the original question need to be |dw:1359975317046:dw|

    • one year ago
  27. Jonask
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    \[\huge x^2=2^{\frac{ 2 }{ 3 }}+2+4^{\frac{ 2 }{ 3 }}\] our expresion \[\huge x^2-4\]

    • one year ago
  28. TeemoTheTerific
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    you lost me there :P

    • one year ago
  29. Jonask
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    oh i found an easier way

    • one year ago
  30. Kuoministers
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    \[\huge (2^{\frac{ 1 }{ 3 }} + 4^{\frac{ 1 }{ 3 }})^{3}\] That is the question...

    • one year ago
  31. Kuoministers
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    a + b)3 = a3 + 3a2b + 3ab2 + b3

    • one year ago
  32. TeemoTheTerific
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    ha i think your right

    • one year ago
  33. Jonask
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    lets factor 2^1/3\[\huge 2^{\frac{ 1 }{ 3 }}(1+2^{\frac{1}{3}})\] \[\huge 2^{\frac{ 1 }{ 3 }3}(1+2^{\frac{ 1 }{ 3 }})^3\] \[\huge 2(1+3(1)2^{1/3}+3(2^{2/3})+2)\] \[\huge 2(3)(1+2^{1/3}+2^{2/3})\] \[\huge 6(1+x)\]

    • one year ago
  34. TeemoTheTerific
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    ok i worked it out by myself but thanks so much for your help jonask

    • one year ago
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