<s>[INST] 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 $
 [/INST] Let $$\sqrt[3]{18 + \sqrt{325}} + \sqrt[3]{18 - \sqrt{325}}=x$$
Thus, $$18 + \sqrt{325}+ 18 - \sqrt{325}+3\left(\sqrt[3]{18 + \sqrt{325}} + \sqrt[3]{18 - \sqrt{325}}\right)\sqrt[3]{18 + \sqrt{325}}\sqrt[3]{18 - \sqrt{325}}=x^3$$ or
$$36+3x\cdot(-1)=x^3$$ or
$$x^3-3x^2+3x^2-9x+12x-36=0$$ or
$$(x-3)(x^2+3x+12)=0.$$
Can you end it now?
Also, by your hint:
$$\sqrt[3]{18 + \sqrt{325}} + \sqrt[3]{18 - \sqrt{325}} =\frac{3}{2}+\frac{1}{2}\sqrt{13}+\frac{3}{2}-\frac{1}{2}\sqrt{13}=3.$$
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