What is the coefficient of $x^{25}$ in $(x^3 + x + 1)^{10}$?

Question

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!

Answer 1

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 $$ N=\binom{10}{8}\binom{2}{1}\binom{1}{1}=90. $$

Answer 2

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