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!

The only way to get $x^{25}$ multiplying $10$ terms among 
$$
x^3,\,\,\,x\,\,\text{and}\,\,1, 
$$
is
$$
25=8\times 3+1\times 1+1\times 0,
$$
i.e.,
$$
x^{25}= (x^3)^8\cdot (x)^1\cdot (1)^1.
$$
[If we we $x^3$ seven times or less, then we get at most $x^{24}$, while if we use $x^3$ nine times or more, then we get at least $x^{27}$. Hence, we need to use $x^3$ eight times, and consequently, $x$ once, and $1$ once.]
Therefore, the coefficient we are looking for (i.e., the # of ways to choose 8 times $x^3$, once $x$ and once $1$) is
$$
N=\binom{10}{8}\binom{2}{1}\binom{1}{1}=90.
$$
