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We need to find a polynomial with integer coefficients whose roots include $x=\sqrt[3]2+\sqrt[3]3$, by hand.

Here are some ideas, but none of them are easy enough by hand.

  1. Write the system of equations$$a^3=2,~b^3=3,~x=a+b.$$Then we can use the resultant to eliminate $a$ and $b$.
  2. Note that for all positive integers $n$, the number $x^n=\left(\sqrt[3]2+\sqrt[3]3\right)^n$ is in the form of $$\sum_{i=0}^2\sum_{j=0}^2c_{ij}\sqrt[3]{2^i\times3^j},$$ where $c_{ij}$ are all integers. Thus, by a linear combination of $1$, $x$, $x^2\dots$, $x^9$, we can eliminate all radicals. But the expansion is already too complicated.
    By the way, the answer given by Mathematica is $x^9-15x^6-87x^3-125$, which means that if we can somehow deduce beforehand that a linear combination of $x^3$, $x^6$, $x^9$ is sufficient, then the work needed for approach 2 is significantly reduced.

So, what is a nice way to solve the problem by hand?

youthdoo
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    The second case isn't so hard if you put $n=3$ – Supernerd411 Jun 10 '25 at 01:43
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    $x^3 = (a+b)^3 = a^3 + b^3 + 3ab(a+b) = 5 + 3abx$, hence $(x^3 - 5)^3 = 27(ab)^3 x^3 = 27 \cdot 6 x^3$. I'm not sure what is computationally difficult about this question? – heropup Jun 10 '25 at 01:46
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    there is a procedure: make one companion matrix $A$ for $x^3 - 2,$ another companion matrix $B$ for $x^3 - 3.$ Then the Kronecker sum of $A$ and $B$ has thecharacteristic polynomial you want – Will Jagy Jun 10 '25 at 02:02

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