Jtdylktuy

How to prove that $left(-18+sqrt{325}right)^{frac{1}{3}}+left(-18-sqrt{325}right)^{frac{1}{3}} = 3$ in a direct way ?

I have found one indirect way to do so: Define $t=left(-18+sqrt{325}right)^{frac{1}{3}}+left(-18-sqrt{325}right)^{frac{1}{3}}$, and then observe $t$ is a root of $f(x) = x^3 + 3x - 36 = 0$. Also observe that $f(3) = 0$. Now $f(x)$ has only one real root, since $f'(x) = x^2+3$ has no real solution. So $t =3$, as desired.

But I couldn't find any direct/algebraic way of fiddling it for a while. I guess morever this is true: $left(-frac{t(t^2+3)}{2}+sqrt{left[frac{t(t^2+3)}{2}right]^2 + 1}right)^{frac{1}{3}}+left(-frac{t(t^2+3)}{2}-sqrt{left[frac{t(t^2+3)}{2}right]^2 + 1}right)^{frac{1}{3}} = t$, which can possibly be proved in the same indirect way but I don't know how to prove this by direct manipulation.

One way maybe is to write $z_+ = left(-frac{t(t^2+3)}{2}+sqrt{left[frac{t(t^2+3)}{2}right]^2 + 1}right)^{frac{1}{3}}$, $z_- = left(-frac{t(t^2+3)}{2}-sqrt{left[frac{t(t^2+3)}{2}right]^2 + 1}right)^{frac{1}{3}}$, and writing $t = z_+ + z_-$, observe $z_+z_- = -1$, so $(x-z_+)(x-z_-) = x^2 - tx -1$, and then with Fundamental Theorem of Symmetric Polynomials, maybe calculating $(z_+^3 - z_-^3)^2$ and $(z_+^3 + z_-^3)$ in terms of $t$ would help ?

Compute
$$left(frac{3+sqrt{13}}2right)^3=18+5sqrt{13}.$$
Therefore
$$sqrt[3]{18+5sqrt{13}}=frac{3+sqrt{13}}2.$$
Similarly
$$sqrt[3]{pm 18pm 5sqrt{13}}=frac{pm 3pmsqrt{13}}2$$
where the signs on both sides correspond. Then
$$sqrt[3]{-18+ 5sqrt{13}}-sqrt[3]{-18-5sqrt{13}}
=frac{-3+sqrt{13}}2+frac{-3-sqrt{13}}2=-3.$$

Let first root be $A$ and second be $B$, then we need to find the value of $A + B = x$

1. $A^3 + B^3 = -36$


2. $AB = -1$

Expand the cube: $A^3 + B^3 = (A+B)*(A^2 - AB + B^2) = (A+B)*((A+B)^2-3AB) = x*(x^2+3) = x^3 + 3x = -36$.

Easy guess $x = -3$

Another approach: assume that you have a cube under the root, then, say, you take the right equation where you factor the minus out:
$18 + 5sqrt{13} = (a+bsqrt{13})^3$

Then you gotta solve the system of equations made from matching the summands with the $sqrt{13}$ and without:

1. $a^3 + 39ab^2 = 18$


2. $3a^2b + 13b^3 = 5$

Wolfram yields $a = 1.5$ and $b = 0.5$, by hand calculations led me nowhere

It helps a lot that we know the answer in advance! Otherwise, some lucky guesswork (or knowledge of algebraic number theory) is required.

One can follow a quite similar line of reasoning to Rafael Bombelli (c.1526-1572). See Bombelli and the invention of complex numbers; or my answer to an earlier question here; or section 14.3, "Cubic Equations", of John Stillwell's book Mathematics and Its History (second edition 2002 - there is a third edition, which I don't have).

We know that if $a$ and $b$ are rational, and $(a + bsqrt{13})^3 = c + dsqrt{13}$, then $(a - bsqrt{13})^3 = c - dsqrt{13}$. So it is a natural idea - and a much less "radical" leap than Bombelli's (if you'll pardon the pun) - to look for rational $a, b$ such that $(a + bsqrt{13})^3 = 18 + 5sqrt{13}$. Then we will have $sqrt[3]{18 + 5sqrt{13}} + sqrt[3]{18 - 5sqrt{13}} = 2a$.

We know right away that $a = frac{3}{2}$. (Honesty compels me to admit that I initially overlooked this obvious fact!) Equating coefficients of the irrational number $sqrt{13}$ in the expanded version of either of the equations defining $a, b$, we have $a^3 + 39ab^2 = 18$, which simplifies to $frac{9}{4} + 39b^2 = 12$. Taking $b > 0$, for definiteness, we have $b = frac{1}{2}$. We now only
need check that $3a^2b + 13b^3 = 5$, i.e. $3a^2 + 13b^2 = 10$, to see that the
defining equations for $a, b$ are indeed satisfied.