How to prove that $(-18+\sqrt{325})^{\frac{1}{3}}+(-18-\sqrt{325})^{\frac{1}{3}} = 3$
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 $$\sqrt[3]{-18+ 5\sqrt{13}}-\sqrt[3]{-18-5\sqrt{13}} =\frac{-3+\sqrt{13}}2+\frac{-3-\sqrt{13}}2=-3.$$
Let first root be $A$ and second be $B$, then we need to find the value of $A + B = x$
- $A^3 + B^3 = -36$
- $AB = -1$
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: $18 + 5\sqrt{13} = (a+b\sqrt{13})^3$
Then you gotta solve the system of equations made from matching the summands with the $\sqrt{13}$ and without:
- $a^3 + 39ab^2 = 18$
- $3a^2b + 13b^3 = 5$
Wolfram yields $a = 1.5$ and $b = 0.5$, by hand calculations led me nowhere