<s>[INST] Expressing a Polynomial as a sum of cube roots of integers How do you prove $x^3-3x^2-6x-4$ has a zero of the form $\sqrt[3]a+\sqrt[3]b+\sqrt[3]c$, for distinct positive integers a,b,c
 [/INST] We have formulas to compute the roots of a cubic polynomial. 
Use them and you get that it has a real solution $\sqrt[3]{3^2}+\sqrt[3]{3}+\sqrt[3]{1}$. 
Now, suppose that we don't know the formulas (But still, we should somehow know how solving cubic goes). We can replace $y=x-1$.
To get $(y+1)^3−3(y+1)^2−6(y+1)−4=y^3-9y+-12$.
We can try to find $y$ in the form $y=a+b$ with $3ab=9$.
Imposing this you get equations for $a^3$ and $b^3$, from where you will get the $a^3=3$ and $b^3=3^2$.
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