How to prove that $(-18+\sqrt{325})^{\frac{1}{3}}+(-18-\sqrt{325})^{\frac{1}{3}} = 3$

Question

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 ?

Answer 1

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

Answer 2

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 $x = -3$

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

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

Answer 3

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 $(a + b\sqrt{13})^3 = c + d\sqrt{13}$, then $(a - b\sqrt{13})^3 = c - d\sqrt{13}$. 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 $(a + b\sqrt{13})^3 = 18 + 5\sqrt{13}$. Then we will have $\sqrt[3]{18 + 5\sqrt{13}} + \sqrt[3]{18 - 5\sqrt{13}} = 2a$.

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.

Answer 4

By direct calculation we have the identity $a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ac).$

Then set $a=\sqrt[3]{-18+\sqrt{325}},b=\sqrt[3]{-18-\sqrt{325}}$ and $c=3$. Note that $ab=-1$ and $a^3+b^3=-36$, so $a^3+b^3+c^3=3abc$.

Since $(a-b)^2+(b-c)^2+(b-c)^2>0\Longleftrightarrow a^2+b^2+c^2-ab-bc-ac>0$ holds for any three distinct real numbers, we have that $a+b+c=0$, which means $$\sqrt[3]{-18+\sqrt{325}}+\sqrt[3]{-18-\sqrt{325}}=-3.$$