Algebraic expression rewritting

[J Goodrich]

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?

 

[nirax]

there is no finite expansion.

 

[J Goodrich]

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

 

[nirax]

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.

 

[HallsofIvy]

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

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" $x^3- mx= n$?

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

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

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

Of course that means that n/2= 1 and that $(n/2)^2+ (m/3)^3= 28/27$ 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 $x^3- mx= x^3+ x= n= 2$. Obviously, x= 1 is a root of that equation and, dividing by x-1, $x^3+ x- 2= (x-1)(x^2+ x+ 2)= 0$. But, by the quadratic formula, $x^2+ x+ 2= 0$ has NO real roots. The only real root is x= 1 and that must be the one given:
$$\sqrt[3]{1 + \sqrt{28/27}} + \sqrt[3]{1 - \sqrt{28/27}}= 1$$

 

[nirax]

good joke Goodrich :)