- #1

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(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?

- Thread starter J Goodrich
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- #1

- 17

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(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

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there is no finite expansion.

- #3

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So [tex]\sqrt[3]{1 + \sqrt{28/27}} + \sqrt[3]{1 - \sqrt{28/27}}[/tex] has no simpler form?

- #4

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- #5

HallsofIvy

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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

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

[tex]\sqrt[3]{1 + \sqrt{28/27}} + \sqrt[3]{1 - \sqrt{28/27}}= 1[/tex]

- #6

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good joke Goodrich :)

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