TITLE: Find coefficient of $x^8$ in $(1-2x+3x^2-4x^3+5x^4-6x^5+7x^6)^6$
QUESTION [11 upvotes]: Find coefficient of $x^8$ in $(1-2x+3x^2-4x^3+5x^4-6x^5+7x^6)^6$
how to do it? I think it should be $3^6$ since $(3x^2)^6=3^6x^8$. (this is false)
Is this true?

REPLY [3 votes]: Although this case allows the use of Taylor series, which might just be a coincidence, its well worth the time to show the more general approach that is the multinomial formula--which is shown in mathematics2x2life's post. I find using the multinomial formula very helpful in trinomial expressions where the usual binomial doesn't directly work. In this case the polynomial is too large to give a name,
$$(7x^6-6x^5+5x^4-4x^3+3x^2-2x+1)^6$$ 
First let me define the variables in the formula,
$$(x_1+x_2+\cdots x_m)^n=\sum_{i,j,k,...}\binom{n}{i,j,k,...}x_1^{i}x_{2}^{j} x_{3}^{k}...$$
In comparison to our expression we know that $n=6$ and as a result $i+j+k+L+m+n+o=6$ at all times. We have seven letters for the seven terms in the expression. We use the $i,j,k...$ to determine the coefficients of several $x^8$ terms. In this particular case, there were $18$ terms. I will explain the first few but state the others sans explanation. 
To start we use the first term, which is $7x^6$. The value of $i$ clearly can't be a $2$ since that will make the term degree greater than $8$ so that leaves only two possibilities $i=0$ and $i=1$, we ignore $i=0$ because it will come up later when the other $j,k,l,...$ take on their values. To understand what combination to pick, you have to understand the implication of the $i$. Each term has a coefficient as such,
$$\frac{n!}{i!\times j!\times k!...o!}x_{1}^{i}x_{2}^{j}...x_{7}^{o}$$
So, to find the coefficients of all $x^8$ terms we have to look at all the combinations of $i,j,k...$ for which a power of 8th comes up and for which $i+j+k+L+m+n+o=6$. It's like trying to solve a simultaneous equation with constraints, although in this case its just one equation with many constraints. If the constant term, $1$, wasn't in the expression the number of possibilities would decrease down to $2$-clearly simplifying the work. 
Now back to $i=1$, if $i=1 \implies (7x^6)^{i=1}$ then we need to multiply it by a term with $x^2$. Looking at the other terms, there are two possibilities,
$$i=1, \ m=1, \ o=4 \implies \frac{6!}{1!\times 1! \times 4!}(7x^6)^{1}\times(3x^2)^1\times 1^4=630x^8 $$
$$i=1, \ n=2, \ o=3 \implies \frac{6!}{1!\times 2! \times 3!}(7x^6)^{1}\times(2x)^2\times 1^3=1680x^8 $$
$630$ and $1680$ are part of the whole coefficient of $x^8$. Now the rest are shown below:
$$j=1, \ L=1, \ o=4 \implies 720x^8 $$
$$j=1, \ m=1, \ n=1, \ o=3 \implies 4320x^8 $$
$$j=1, \ n=3, \ o=2 \implies 2880x^8 $$
$$k=2, \ o=4 \implies 375x^8 $$
$$k=1, \ L=1, \ n=1, \ o=3 \implies 4800x^8 $$
$$k=1, \ m=2, \ o=3 \implies 2700x^8 $$
$$k=1, \ n=4, \ o=1 \implies 2400x^8 $$
$$k=1, \ m=1, \ n=2, \ o=2 \implies 10800x^8 $$
$$L=2, \ m=1, \ o=3 \implies 2880x^8 $$
$$L=2, \ n=2, \ o=2 \implies 5760x^8 $$
$$L=1, \ m=2, \ n=1, \ o=2 \implies 12960x^8 $$
$$L=1, \ m=1, \ n=3, \ o=1 \implies 11520x^8 $$
$$L=1, \ n=5, \implies 768x^8 $$
$$m=4, o=2 \implies 1215x^8 $$
$$m=3, \ n=2, \ o=1 \implies 6480x^8 $$
$$m=2, \ n=4 \implies 2160x^8 $$
Note that I haven't stated the missing $i,j,k,...$ since they are zero. Now to find the final coefficient, we add all the individual terms to get $75048x^8.$