This is a classic example of binomial expansion.
A few simple expansion of (a+b) to different powers:
(a+b)^2=a^2+2ab+b^2
(a+b)^3=a^3+3a^2b+3ab^2+b^3
(a+b)^4=a^4+4a^3b+6a^2b^2+4ab^3+b^4
You may begin to see a pattern in the coefficients. They come from Pascal's Triangle:
" 1"
" 1 1"
" 1 2 1 "color(grey)((a+b)^2)
" 1 3 3 1 "color(grey)((a+b)^3)
" 1 4 6 4 1 "color(grey)((a+b)^4)
"1 5 10 10 5 1"
...
Each number is the sum of the two numbers above it. The entries are also equal to ""_nC_k, where n is the row number and k is the (diagonal) column number. (The first row is "row 0", and the first column is "column 0".)
To expand a general binomial like (a+b)^n, we use the following form:
(a+b)^n=""_nC_0a^n+""_nC_1a^(n-1)b^1+""_nC_2a^(n-2)b^2+...+""_nC_nb^n
As the terms go by, the a exponent counts down from n, while the b exponent counts up from 0. So the term with a^k in it will be:
color(white)1_n C_(n-k)a^kb^(n-k).
In this question, n=9, k=6, a=x, and b="-"3. Thus, the term with x^6 in it is
color(white)1_9C_(9-6)x^6("-"3)^(9-6)
=""_9C_3x^6("-"3)^3
=84x^6("-"27)
=-2268x^6
So the coefficient is -2268.
Bonus:
We can generalize this to find the x^k term in a general binomial (ax+by)^n, where a and b are numbers. The formula will be:
(color(white)1_nC_k*a^k*b^(n-k))x^ky^(n-k)
and the product in the brackets will be our coefficient. For example, in the above problem, we have a=1,b="-3" (and y=1). When n=9 and k=6, this gives:
color(white)1_9C_6*(1)^6*("-3")^(9-6)
which is the same value as before. (Note: color(white)1_nC_k=color(white)1_nC_(n-k).)
