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No idea -- algebraic manipulation involving powers

  1. May 31, 2016 #1
    1. The problem statement, all variables and given/known data
    If x=2+2^1/2+2^2/3. Then x^3-6x^2+6x=?

    2. Relevant equations
    (A+b+c)^2=a^2+b^2+c^2+2(ab+bc+ca+)

    3. The attempt at a solution
    x^2= 6+3(2^2/3)+2^4/3+4(2^1/3)
     
  2. jcsd
  3. May 31, 2016 #2
    Should I go on and find x^3
     
  4. May 31, 2016 #3
    Hint :- take '2' to the left side and then cube both sides.
     
    Last edited: May 31, 2016
  5. May 31, 2016 #4
  6. May 31, 2016 #5
    Got x^3-6x^2+12x=14+6(2^1/3+2^2/3)
     
  7. May 31, 2016 #6
    I think there is a calculation mistake here, sorry if i am wrong.
     
    Last edited: May 31, 2016
  8. May 31, 2016 #7

    Nidum

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    Work out the numerical value of x from the first equation .

    What do you do then ?
     
    Last edited: May 31, 2016
  9. May 31, 2016 #8
    The question is solvable without calculating square roots and cube roots.
     
  10. May 31, 2016 #9
    factorize the polynomial, if all go well (i.e the polynomial has 3 real roots) you should get it in the form ##(x-r_1)(x-r_2)(x-r_3)##

    You can easily see that one root of the polynomial is....
     
    Last edited: May 31, 2016
  11. May 31, 2016 #10

    Nidum

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    Think about what sort of answer the question really requires .
     
  12. May 31, 2016 #11
    That might just do it
     
  13. May 31, 2016 #12
    Which one should I factorise
     
  14. May 31, 2016 #13
    ##x^3-6x^2+6x=x(x^2-...+...)## the polynomial inside the parenthesis is a 2nd order polynomial i believe you know how to factorize it,
     
  15. May 31, 2016 #14
    Wait
    I could just substitute then
    X^3-6x^2+12x=14+6(2^1/3+2^2/3). ==:
    x^3-6x^2+12x=14+6x-12. ==:
    Ans. Is 2
     
  16. May 31, 2016 #15
    Thank You All!!!!!!!
     
  17. May 31, 2016 #16
    Your question is wrong. it should be :- x= 2 + 2^(1/3) + 2^(2/3)

    otherwise your calculations are wrong, you can just check.
     
  18. May 31, 2016 #17

    SammyS

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    Hello Chaos_Enlightened. Welcome to PF !

    Your original expression, has some ambiguity in it.

    Using the standard Order of Operations, What you wrote, x=2+2^1/2+2^2/3 , literally means ## \ x=2+2^1/2+2^2/3\ .##

    Use parentheses around the fractional exponents as in x=2+2^(1/2)+2^(2/3), which gives ## \ x=2+2^{1/2}+2^{2/3}\ .##

    You may find the superscript feature (X2 button) helpful as well. Using this gives: x=2+21/2+22/3.

    LaTeX was used for these.
     
  19. May 31, 2016 #18
    Thank you Sammy's
     
  20. May 31, 2016 #19
    I dont think that it factorizes any more than that at least in integers. I could be wrong though it would not be the first time...

    I think it seems that the first common and therefore true factor is the lonely (x) (something...)

    (x) (x2 -6x +6)

    closest I managed to get was (x-3)(x-3) which yields out too much stuff. x^2 -6x + 9

    (x-2) (x-3) yields too little stuff x^2 -5x + 6

    x* x * (x-6 + 6/x) = (x^3) - (6x^2) +( 6x)
     
  21. Jun 1, 2016 #20
    for x2 -6x +6 = 0

    x is either (6+ √12) / 2
    or
    x is (6-√12) / 2

    Its not very easy to calculate the original equation when the tip is used to switch 2 to the left side and cube both sides. Newton's binomial formula or pascals triangle can be used here. Maybe I am on the wrong track but looking at the calculation it will not be easy or fast to compute those roots into cubes. (X-2)^3 will be easier but also quite tedious.
    x= 2 + 20.5 + 22/3
    (x-2)3 = [(√2) + 3√4) ] 3


    (x-2)^3 = [x+(-2)]3
    calculate with binomial theorem or Newton's binomial formula which is found in some mathematics formula books ( It seems I used simplified version from wikipedia https://en.wikipedia.org/wiki/Binomial_theorem scroll down in the article until there is (x+y)3). At least the factors themselves can be seen from pascal's triangle easily. In this case the factors are 1,3,3, and 1.

    x3-6x2 +12x -8

    I'm sure that the right hand side yields something even uglier with pen-and--paper calculation... ?:)
     
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