<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] Since $(3\pm\sqrt{13})^3=144\pm40\sqrt{13}=8(18\pm5\sqrt{13}),$
$\sqrt[3]{18\pm\sqrt{325}}=\sqrt[3]{18\pm5\sqrt{13}}=\dfrac{3\pm\sqrt{13}}2$.
Therefore, $\sqrt[3]{18+\sqrt{325}}+\sqrt[3]{18-\sqrt{325}}=\dfrac{3+\sqrt{13}}2+\dfrac{3-\sqrt{13}}2=3.$
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