10.1. Having found some polynomial of the sixth degree $x^{6}+a_{1} x^{5}+\ldots+a_{5} x+a_{6}$ with integer coefficients, one of the roots of which is the number $\sqrt{2}+\sqrt[3]{5}$, write in the answer the sum of its coefficients $a_{1}+a_{2}+\ldots+a_{6}$.
Answer: -47. Solution. $x=\sqrt{2}+\sqrt[3]{5} \Rightarrow (x-\sqrt{2})^{3}=5 \Rightarrow x^{3}-3 x^{2} \sqrt{2}+3 x \cdot 2-2 \sqrt{2}=5$ $\Rightarrow x^{3}+6 x-5=(3 x^{2}+2) \cdot \sqrt{2} \Rightarrow (x^{3}+6 x-5)^{2}=(3 x^{2}+2)^{2} \cdot 2$ $\Rightarrow P_{6}(x)=(x^{3}+6 x-5)^{2}-2(3 x^{2}+2)^{2}$. In this case, the polynomial itself does not need to be explicitly obtained (for reference, note that this is the polynomial $x^{6}-6 x^{4}-10 x^{3}+12 x^{2}-60 x+17$), since we need the sum: $a_{1}+\ldots+a_{6}=P_{6}(1)-1=2^{2}-2 \cdot 5^{2}-1=-47$.