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Algebraic expression rewritting

  1. Aug 27, 2009 #1
    is there any expansion of

    (a+b)^(1/3) + (a-b)^(1/3)

    or any other manipulation that could be done to rewrite this expression, perhaps under one radical or in some other way rather than the addition of two radicals?
  2. jcsd
  3. Aug 27, 2009 #2
    there is no finite expansion.
  4. Aug 29, 2009 #3
    So [tex]\sqrt[3]{1 + \sqrt{28/27}} + \sqrt[3]{1 - \sqrt{28/27}}[/tex] has no simpler form?
  5. Aug 29, 2009 #4
    no. it is a nonrepeating real number and and the expression you wrote is the exact one. most of the (algebraic) numbers are like that only.
  6. Aug 30, 2009 #5


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    Actually, there is a very easy way of writing that: it is 1!

    You could, of course, do that computation on a calculator and get the result "1" but that would only show that it is "approximately" 1. I mean that it is exactly 1.

    I noticed that this looks a lot like the form of solution one would expect from Cardano's cubic formula. The basic idea goes like this:

    Let x= a+ b. Then [itex]x^3= a^3+ 3a^2b+ 3ab^2+ b^3[/itex]. Also, [itex]-3abx= -3ab(a+ b)= -3a^2b- 3ab^2[/itex] so that [itex]x^3- 3abx= a^3+ b^3[/itex]. Now let m= 3ab and [itex]n= a^3+ b^3[/itex] and we have shown that x satisfies [itex]x^3- mx= n[/itex].

    Now, what about the other way around? Suppose we are given m and n. Can we find a and b and so find x= a+ b satisfying the "reduced cubic" [itex]x^3- mx= n[/itex]?

    Yes, we can! From m= 3ab, b= m/3a. Then [itex]n= a^3- b^3= a^3- m^3/(3^3a^3)[/itex] and, multiplying by [itex]a^3[/itex], we get the quadratic, in [itex]a^3[/itex], [itex]na^3= (a^3)^2- (m/3)^3[/itex] or [itex](a^3)^2- n(a^3)- (m/3)^3[/itex]. By the quadratic formula, that has solution
    [tex]a^3= \frac{n\pm\sqrt{n^2- 4(m/3)^3}}{2}= \frac{n}{2}\pm\sqrt{(n/2)^2- (m/3)^3z}[/tex]

    If we take the positive sign, then, from [itex]b^3= n- a^3[/itex], we have
    [tex]b^3= \frac{n}{2}- \sqrt{(n/2)^3- (m/3)^3}[/tex] and x= a+ b.

    Now doesn't
    [tex]\sqrt[3]{1+ \sqrt{\frac{28}{27}}+\sqrt[3]{1- \sqrt{\frac{28}{27}}[/tex]
    look exactly like that?

    Of course that means that n/2= 1 and that [itex](n/2)^2+ (m/3)^3= 28/27[/itex] Putting n/2= 1 into the second of those, (m/3)^3= 28/27- 1= -1/27 and m/3= -1/3 so m= -1. That means that the given number is a real root of [itex]x^3- mx= x^3+ x= n= 2[/itex]. Obviously, x= 1 is a root of that equation and, dividing by x-1, [itex]x^3+ x- 2= (x-1)(x^2+ x+ 2)= 0[/itex]. But, by the quadratic formula, [itex]x^2+ x+ 2= 0[/itex] has NO real roots. The only real root is x= 1 and that must be the one given:
    [tex]\sqrt[3]{1 + \sqrt{28/27}} + \sqrt[3]{1 - \sqrt{28/27}}= 1[/tex]
  7. Aug 31, 2009 #6
    good joke Goodrich :)
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