Don't solve this equation, please!

Relevant wiki: Vieta's Formula Problem Solving - Intermediate

Let the roots of the equation $x^9+m_1x^8+m_2x^7+m_3x^6+m_4x^5+m_5x^4+m_6x^3+m_7x^2+m_8x+m_9=0$ be $a_1,a_2,a_3...a_9$ .

Then $a_1a_2a_3...a_9=-m_9$ ,and $a_1^9+a_2^9+a_3^9+...+a_9^9=f_9(m_1,m_2,m_3,...m_8)+nm_9$ where $f_9$ is some function,and $n$ is a constant.

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Let's find $n$ first.Consider the equations $x^9=0$ and $x^9-1=0$ .

The first equation's roots are $b_1,b_2,b_3,...b_9$ ,and the second one's are $c_1,c_2,c_3,...c_9$

According to the above,we have $b_1^9+b_2^9+b_3^9+...+b_9^9=0=f_9(0,0,0,...,0)+0n$ and $c_1^9+c_2^9+c_3^9+...+c_9^9=9=f_9(0,0,0,...,0)-1n$

So we get $n=-9$

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Consider the 2 equations $\left(x-\sqrt[9]{2}\right)\left(x-\sqrt[9]{3}\right)\left(x-\sqrt[9]{4}\right)\ldots\left(x-\sqrt[9]{10}\right)-1=0$ and $\left(x-\sqrt[9]{2}\right)\left(x-\sqrt[9]{3}\right)\left(x-\sqrt[9]{4}\right)\ldots\left(x-\sqrt[9]{10}\right)=0$ Expand them,and they will be in the form of $x^9+p_1x^8+p_2x^7+p_3x^6+p_4x^5+p_5x^4+p_6x^3+p_7x^2+p_8x-\sqrt[9]{10!}-1=0$ and $x^9+p_1x^8+p_2x^7+p_3x^6+p_4x^5+p_5x^4+p_6x^3+p_7x^2+p_8x-\sqrt[9]{10!}=0$ where $p_1,p_2,...,p_8$ are constants.

According to the above,we have $x_1^9+x_2^9+x_3^9+...+x_9^9=f_9(p_1,p_2,p_3,...,p_8)+9(\sqrt[9]{10!}+1)$ and $2+3+4+...+10=54=f_9(p_1,p_2,p_3,...,p_8)+9\sqrt[9]{10!}$

Subtract and get $x_1^9+x_2^9+x_3^9+...+x_9^9=\boxed{63}$

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To get a clearer idea about expressing $a_1^9+a_2^9+a_3^9+...+a_9^9$ in the form of $f_9(m_1,m_2,m_3,...m_8)+nm_9$ ,here is an example of a third degree one.

If the equation $x^3+px^2+qx+r=0$ has roots $a,b,c$ ,then

$a^3+b^3+c^3=(a+b+c)^3-3(a+b+c)(ab+bc+ca)+3abc=-p^3+3pq-3r$ If $f_3(x,y)=-x^3+3xy,m=-3$ ,then $a^3+b^3+c^3$ can be expressed in the form of $f_3(p,q)+mr$