What is the coefficient of $x^{25}$ in $(x^3 + x + 1)^{10}$? Here's what I have so far on the off chance that my thinking is correct...
So using Vieta's the coefficient of the $x^{25}$ should be $-(r_1r_2r_3r_4r_5+r_1r_3r_4r_5r_6+...+r_6r_7r_8r_9r_{10})$
Since each root is equal it is $-(r^5+r^5...+r^5)$
And there are 10 roots and you want to group them in 5's, so there are $10 \choose 5$ = 252 number of $r^5$s.
Have I gotten it right so far or are there errors in my work and/or line of reasoning?
I have $-252*(r^5)$ as the coefficient, with r being the solution to the cubic $x^3+x+1=0$.
How do I solve for r?
Would appreciate any help with this problem. Thanks in advance!

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With $\ds{a,b,c\in{\mathbb Z}}$:

\begin{align}&\color{#c00000}{\pars{x^{3} + x + 1}^{10}}
=\sum_{%
\vphantom{\huge A}a,b,c\ \geq\ 0\atop{\vphantom{\LARGE A}a\ +\ b\ +\ c\ =\ 10}}
{10! \over a!\ b!\ c!}\,\pars{x^{3}}^{a}x^{b}1^{c}
=\sum_{%
\vphantom{\huge A}a,b,c\ \geq\ 0\atop{\vphantom{\LARGE A}a\ +\ b\ +\ c\ =\ 10}}
{10! \over a!\ b!\ c!}\,x^{3a + b}
\end{align}

With $\ds{3a + b = 25}$ and $\ds{a + b + c = 10}$ we'll have
  $\ds{a = {15 + c \over 2}}$ and $\ds{b = {5 - 3c \over 2}}$. Also,
  $\ds{0 \leq b\ \imp\ c \leq {5 \over 3}\ \imp\ c = 1}$ because $\ds{a}$ is an integer ( $\ds{c}$ must be odd ).

We get the $\ds{x^{25}}$-coefficient with
$\ds{\quad a = {15 + \color{#c00000}{1} \over 2} = 8\,,\quad
b = {5 - 3\times\color{#c00000}{1} \over 2} =1\,,\quad c = \color{#c00000}{1}}$:
$$
{10! \over 8!\ 1!\ 1!}=10\times 9=\color{#66f}{\Large 90}
$$
