<s>[INST] Let $a\gt 2 $ be a root of the equation $x^3-x-8 = 0$. Compute $\sqrt[3]{6a^2-13a} + \sqrt[3]{6a^2+13a+16}$ I came across this question in some past contest prep papers belonging to Akdeniz University. To solve it, I've tried letting the cuberoots equal some $k$ and $l$ and tried writing $k+l$ in terms of $kl$ and $k^3+l^3$ but to no avail. Any help would be appreciated.
 [/INST] Using $a^3-a-8=0$ one gets:
$$(2+a)^3=a^3+6a^2+12a+8=(a^3-a-8)+6a^2+13a+16=6a^2+13a+16$$
$$(2-a)^3=-a^3+6a^2-12a+8=-(a^3-a-8)+6a^2-13a=6a^2-13a$$
So $\sqrt[3]{6a^2+13a+16}+\sqrt[3]{6a^2-13a}=2+a+2-a=4$
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