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Homework Help: 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

    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


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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


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    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


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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


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    ##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
    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


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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
    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... ?:)
  22. Jun 1, 2016 #21
    I think I got the answer by looking at what I got from the binomial formula.

    • I recommend you start from the left side first (x-2)3

    • I tend to agree with Sahil Kukreja's advice. He or she provided a good tip. ( I think ) .
    • I think I got the basic idea, in order to get the "rough answer" in unfinished form. It should still be mathematically correct but your teacher will likely have the finished form answer in the answerbook
    • try to get the "rough form answer" first, then worry later about expanding the (squareroot + cubic root)3
    • Mathematically it's not very significant - it's more of a style issue, I suppose.

    (x-2) 3 = (20.5 + 22/3) 3

    . In the end phase we must give the answer for the question "What does this clause yield: clause is; x3 -6x2 + 6x"
    The answer must be another clause, which equals that quadratic clause. The full clause in the ending (both sides of equation) will be an equation, in that sense. So, therefore the ending clause should be properly balanced like all equations, if that makes it clear. This is my interpretation at least.
    But the question asks simply (what does the left side clause give you, on the right side clause?)

    • We know for a fact that this equation holds true
    x=2 + √2 + 3√4

    • Therefore
    (x-2)3 = [√2 + 3√4] 3 This equation must hold true as well (it f*****g should hold true!!!)

    • Therefore
    What does the clause X3 -6X2 +6X equal to?
    Give the clause, which equals to that earlier clause. [ X3 -6X2 +6X ]


    1. first equation was x= 2 + 20.5 + 22/3
    2. in the first equation, minus 2 from both sides.
    3. in the first equation, then cube both sides (third power both sides)
    4. calculate the (x-2)3 with Newton's binomial formula, or use the wikipedia article about binomial theorem, and find the formula for (x+y)3. Use pen-and-paper and be careful with calculation!

    5. according to my limited knowledge, you can substitute the negative sign, simply by using brackets and minus inside
    (x-2)^3 = [x+(-2)] ^3

    When you calculate
    it is the same as

    At least I hope so because otherwise I screwed up my own solution hehe... :oops:
  23. Jun 1, 2016 #22
    It appears my method yields correct result. I did the rough answer first on pen-and-paper.

    Then I plugged my paper results into the calculator and verified the equation holds true on both sides.
    I think that it may be unnecessary to expand the (20.5+22/3) 3, especially if you have calculator available to verify.

    But one method is to use the binomial formula for (x+y)3. I can't think of any easier way to tackle this problem then simply expand that one out... until you have some result according to the formula.

    x3 -6x2 + 6x = ?

    Calculator can be used at the end to verify that both sides of equation hold true. That's what I would do, but I could be wrong though.
  24. Jun 1, 2016 #23


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    Where did you get ##\ 6\left(2^{1/3}+2^{2/3}\right) \ ?##

    ... or should the original problem state that ##\ x=2+2^{1/3}+2^{2/3}\ ## rather than ##\ x=2+2^{1/2}+2^{2/3}\ ? ##
  25. Jun 1, 2016 #24
    Yes, thats what i was saying in post #16.
  26. Jun 1, 2016 #25


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    Post #16:
    I see that I missed that post of yours.

    Then it's questionable as to whether OP will return to tell us. He's had over a day to respond to your post.

    Certainly, ##\ x=2+2^{1/3}+2^{2/3}\ ## is much easier to work with. However, I was surprised that the problem as posted (likely with a typo) gave a fairly simple result, having only two terms.

    ##\ 6\cdot 2^{5/6}-4\cdot2^{1/2}=5.033\dots\ ##
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