<s>[INST] Why is $\sqrt[3]{18+5\sqrt{13}} + \sqrt[3]{18-5\sqrt{13}} = 3$? 
How to show that $$\sqrt[3]{18+5\sqrt{13}} + \sqrt[3]{18-5\sqrt{13}} = 3?$$

This equality comes from solving $$t^3 - 15 t - 4 = 0$$ using Cardanos fomula and knowing the solution $t_1=4$.
I have attempted multiplying the whole thing with $(\sqrt[3]{18+5\sqrt{13}})^2 - (\sqrt[3]{18-5\sqrt{13}})^2$, but no success. Then I have solved for one cubic root and put all to the third power. Also no success.
 [/INST] $$\left(\sqrt[3]{18+5\sqrt{13}} + \sqrt[3]{18-5\sqrt{13}}\right)^3=$$
$$=18+5\sqrt{13}+18-5\sqrt{13}+3\sqrt[3]{18+5\sqrt{13}} \sqrt[3]{18-5\sqrt{13}}\left(\sqrt[3]{18+5\sqrt{13}} + \sqrt[3]{18-5\sqrt{13}}\right)=$$
$$=36+3\sqrt[3]{-1}\left(\sqrt[3]{18+5\sqrt{13}} + \sqrt[3]{18-5\sqrt{13}}\right)=$$
$$=36-3\left(\sqrt[3]{18+5\sqrt{13}} + \sqrt[3]{18-5\sqrt{13}}\right).$$
Now, let $\sqrt[3]{18+5\sqrt{13}} + \sqrt[3]{18-5\sqrt{13}}=x.$
Thus, $$x^3=36-3x$$ or
$$x^3-3x^2+3x^2-9x+12x-36=0$$ or
$$(x-3)(x^2+3x+12)=0,$$ which gives $x=3$.
But I think the best way in this formulation it's the way by using
$$a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-ac-bc).$$
Now, we need to prove that
$$18+5\sqrt{13}+18-5\sqrt{13}-27-3\cdot(-3)\sqrt[3]{18+5\sqrt{13}}\cdot\sqrt[3]{18-5\sqrt{13}}=0$$ or
$$36-27-3(-3)(-1)=0,$$
which is true.
Done!
</s>