Prove that $\sqrt[3]{18 + \sqrt{325}} + \sqrt[3]{18 - \sqrt{325}} = 3$ without using Cardano's formula. (Hint, what is $(3\pm \sqrt{13})^3$ Prove that $\sqrt[3]{18 + \sqrt{325}} + \sqrt[3]{18 - \sqrt{325}} = 3$ without using Cardano's formula. (Hint, what is $(3\pm \sqrt{13})^3$
I have that $$(3 + \sqrt{13})^3 = 144  + 40 \sqrt{13} $$ and $$(3 - \sqrt{13})^3 = 144  - 40 \sqrt{13} $$
A cursory look into Bombelli's method led me to  the following system of equations: 
$$\sqrt[3]{18 + \sqrt{325}} = a + b^{1/2}$$
$$\sqrt[3]{18 - \sqrt{325}} = a - b^{1/2}$$
I am unsure how to solve this system of equations without making a mess of the radicals...I know however that the given cube roots on the LHS of the above system are solutions to the cubic $x^3 + 3x = 36 $

Hint, a link to a nice resource and to complement Michael's answer. There is this famous statement

If $a+b+c=0$ then $a^3+b^3+c^3=3abc$

source, Mathematical Olympiad Treasures, by Titu Andreescu and Bogdan Enescu, first chapter is available in preview, the proof is there.

As a result 
$$x=\sqrt[3]{18 + \sqrt{325}} + \sqrt[3]{18 - \sqrt{325}} \Rightarrow \\
x-\sqrt[3]{18 + \sqrt{325}} - \sqrt[3]{18 - \sqrt{325}} =0 \Rightarrow \\
x^3- (18 + \sqrt{325}) - (18 - \sqrt{325}) = 3x\sqrt[3]{(18 + \sqrt{325})(18 - \sqrt{325})} \Rightarrow \\
x^3-36=-3x$$
which has $3$ as the only real solution.
