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
