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TeemoTheTerific

  • 2 years ago

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

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  1. Razzputin
    • 2 years ago
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    hello

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

  3. TeemoTheTerific
    • 2 years ago
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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)

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

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

  6. Razzputin
    • 2 years ago
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    have you got a scanner?

  7. TeemoTheTerific
    • 2 years ago
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    no...

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

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

  10. TeemoTheTerific
    • 2 years ago
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    |dw:1359973881754:dw|

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

  12. TeemoTheTerific
    • 2 years ago
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    |dw:1359973981397:dw|

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

  14. Razzputin
    • 2 years ago
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    do you get x^3 = 23.08393261?

  15. TeemoTheTerific
    • 2 years ago
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    im pretty sure we need to show it using indices

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

  17. TeemoTheTerific
    • 2 years ago
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    well i now completely lost with this quesiton :P

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

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

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

  21. Razzputin
    • 2 years ago
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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?

  22. Jonask
    • 2 years ago
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    we not meant to find x but prove

  23. Razzputin
    • 2 years ago
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    its basic substitution

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

  25. Jonask
    • 2 years ago
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    yes

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

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

  28. TeemoTheTerific
    • 2 years ago
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    you lost me there :P

  29. Jonask
    • 2 years ago
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    oh i found an easier way

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

  31. Kuoministers
    • 2 years ago
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    a + b)3 = a3 + 3a2b + 3ab2 + b3

  32. TeemoTheTerific
    • 2 years ago
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    ha i think your right

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

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

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