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Quick Cubic Problem

  1. Aug 1, 2004 #1
    Is anybody interested in this problem? Solve This One:

    Victor Niederhoffer in "Education of a Speculator," recalls that he wanted to join his Brooklyn High School Math Team, and prepair for a hope-for entrance to Harvard University, which he achieved. He mentions that he was asked to solve in about a minute this problem,

    (2+(5)^.5)^1/3 + (2-(5)^.5)^1/3 = ?

    Niederhoffer knew he could not do it, but since it was a multiple choice question he eliminated answers containing pi and e, and guessed the right remaining one.

    He once sent me an autographed copy of his book, and I have always enjoyed this math question. Does anyone know the answer?
  2. jcsd
  3. Aug 1, 2004 #2


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    If 1 is one of the choices, I'm guessing that. Actually getting there in a way that works regardless of the numbers and without a lot of work is going to take a little thought - at least more than a minute's worth.

    Edit: Or negative one (taking a second thought about it). Hoping all of the secondary products cancel out, but not sure - Definitely less than 2 and greater than negative 2.
    Last edited: Aug 1, 2004
  4. Aug 1, 2004 #3
    I'm guessing that without using a calculator the fastest way to solve this type of problem is just to use some simple quick reasonsing. The first part cubrt(2+sqr(5)) you know must be equal to 1.6...... something. Think about it the sqrt(5)= 2.2....something, so you have now cubert(4.2) which is roughly 1.6 something. Using some simple reasoning you will see that you don't have to solve the other part cubrt(2-sqrt(5)). It will just be the .6 something of the first part except with a negative sign. So the answer must be 1. I think to solve this problem in less than a minute you have to be able to do very rough arithmetic very fast.
  5. Aug 1, 2004 #4


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    Knowing the choices would greatly help. :smile:

    The number itself looks as if it was derived via the cubic formula; if you happened to know the cubic formula by heart, you could quickly reconstruct the cubic equation, and then simply test the choices to see which one is a root of the cubic. (Of course, you might have the annoying possibility that two are!)

    Without knowing the cubic, you could still reconstruct the polynomial, but it takes a bit more work.

    I certainly agree, however, that the answer must be near one.
  6. Aug 1, 2004 #5
    "1" was one of the answers, I just did not want to give it away. (I don't remember the 4th choice). It could be a very sticky problem, Niederhoffer never suggested he could have actually worked out the details in a minute.

    However, it can be approached--at least at home-- by setting it equal to Y and then cubing. We assume the principal cubic root, giving (-1)^1/3 = -1.

    We have then Y^3 = 2+5^.5 +2-5^.5 -3{(2+5^.5)^1/3 + (2-5^1/3)} =4-3Y.

    Reinserting the Y, rather than attempting to resolve surds seems the way to go. If we proceed further, we can divide out the Y-1, and get the equation: Y^2+Y+4, which resolves in terms of the square root of -15. But, assumedly, nothing as complicated as that was among the possible answers.
    Last edited: Aug 1, 2004
  7. Aug 1, 2004 #6


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    The two numbers being added are the golden ratio, (1+sqrt(5))/2 and 1 minus the golden ratio (1-sqrt(5))/2, you can verify this easily by cubing. Guessing this was the case is a little trickier, the sqrt(5) pointed that direction as did the fact that 2+sqrt(5)=-1/(2-sqrt(5)).
  8. Aug 2, 2004 #7


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    (2+(5)^.5)^1/3 + (2-(5)^.5)^1/3 = ?

    Take the cube of th expression:

    (2+(5)^.5)^1/3 + (2-(5)^.5)^1/3 = X


    There is one solution, X=1

  9. Aug 2, 2004 #8
    There is a mistake there since (X-1)(X^2+X+1) = X^3 -1, and that gives all the roots of unity.
  10. Aug 2, 2004 #9
    The "2" does not extend across the expression since,

    [2+sqrt(5)]^1/3 = (2)^1/3*[1+(1/2)sqrt(5)]^1/3.
    Last edited: Aug 2, 2004
  11. Aug 2, 2004 #10


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    That's fine...


    so [tex]\frac{1+\sqrt{5}}{2}=(2+\sqrt{5})^{1/3}[/tex].
  12. Aug 3, 2004 #11
    You have made a wonderful observation that makes the problem much simpler, and offers the possibility of an immediate deduction. And I wish I could just understand how to write the math symbols so well also. But I do not agree that there is no other solution, and if I could math-type better here I could more easily show my reasoning to you. You have shown that:

    [(1+sqrt(5))/2)]^3 = (2+sqrt(5)), which is extremely helpful, and that for its "conjugate" we have the answer (2-sqrt(5)). But when we take a cube root we would get three answers. Let w stand for the cubic root of 1, then X, Xw, and Xw^2 are all answers. Furthermore, to make the math simpler, let g stand for (1+sqrt5)/2 and g* stand for (1-sqrt5)/2.

    Now if we "mix up" w and w^2, we get the answer: gw+g*w^2. This I believe is equal to (-1+sqrt(15)i)/2, which comes from solving the remaining quadratic from the equation: Y^3 + 3Y -4.

    The real part of gw+g*w^2 is going to be (w+w^2)/2, but since 1+w+w^2 = 0, we have (w+w^2)/2 = -1/2. The more difficult part is going to be:

    (1/2)sqrt(5)[w-w^2]. w = 1/2(-1+sqrt(3)i), and w^2 =1/2(-1-sqrt(3)i);

    so that w-w^2 = sqrt(3)i, thus we get the product (1/2)sqrt(15)i.

    However, I am not certain about all the answers, but normally in solving the cubic we do, it seems, "mix up" the w and the w^2, when we are looking for more than one answer. The "other way" we would have gotten w and w^2 along with 1 also as roots.

    Let us then return to the equation Y^3 + 3Y-4 = 0, we know that the sum of the roots should be the Y^2 term or zero, while the product of the roots is minus the constant term or 4. Roots such as 1,w,w^2 have a product that is just 1, and so could not qualify, but 1, (-1+sqrt(15)i)/2 and (-1-sqrt(15)i)/2 satisfy both conditions.
    Last edited: Aug 3, 2004
  13. Aug 3, 2004 #12


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    Well, sorry, I've made a mistake,
    [tex]X^3-1+3X-3=(X-1)(X^2+X+1)+3(X-1)=0[/tex], [tex](X-1)[/tex] is factored out, and the second factor is [tex](X^2+X+4)[/tex]. There is one real root, unity, and other two complex ones, those, you wrote[tex]0.5(-1\pm\i\sqrt{15})[/tex]. But I don't think the exeminers wanted complex values.

    I never believed that I would be able to use TEX :)

  14. Aug 3, 2004 #13
    I did not think they wanted complex answers either, and wanted to let that sleeping dog lie.
  15. Aug 3, 2004 #14


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    Of course you're right that there are three answers to [tex](2+\sqrt{5})^{1/3}[/tex], if you allow complex numbers. As you pointed out 2 are complex, one is real, the real one being the golden ratio. I'm 100% confidant that the question is looking for the sum of the real roots, otherwise the problem is not well defined and would have several different answers depending on which roots you took for the respective cube roots.

    ps. take a look at https://www.physicsforums.com/misc/howtolatex.pdf for a quick latex guide. You can also click directly on the latex'ed portions of someones post to see the code they used to generate it.
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