<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] Use a classic method for solving a cubic (not having to remember the formula).
First eliminate the squared term by setting $y=x-1$ to obtain $$y^3-9y-12=0$$
Now we look for solutions of the form $y=v+w$, where $y^3=(v+w)^3=(v^3+w^3)+3vw(v+w)$ which we can rewrite as $$y^3-3vw\cdot y-(v^3+w^3)=0$$
Comparing coefficients we obtain: $3vw=9$ from which we derive $$v^3w^3=27$$ and we also have $$v^3+w^3=12$$ which means that $v^3, w^3$ are the roots of the quadratic $$z^2-12z+27=0=(z-3)(z-9)$$[or use the quadratic formula]
So $v^3=3, w^3=9$ (or the other way about). $y=v+w$, and $x=v+w+1$.
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