Answer:
145,152
Step-by-step explanation:
So there's actually two things you'll need in this equation. You'll need to use pascals triangle, to find the coefficients, and the binomial theorem to find the degrees.
So in this case you want to find the 9th row of pascals triangle. You could write out the entire 10 rows of pascals triangle to find the 9th row (because the first row is row 0). So to find the nth row you generally will start with 1 as the first term as every single row will start with this number. Now after that you're going to continue "row" amount of times starting with 2 values. One value which I'll name k=0 and the other j=1
Now let's start with what we have {1}, take the previous term and (which will always be 1 in the first case, and then multiply it by (row - k) / j. So in this case k=0, and j=1, row=9. You'll have (1 * (9)) / 1. This gives you 9, which is the second term. So now you have the terms {1, 9}. Now the next iteration add 1 to k and 1 to j. And the previous is now 9. So now you have (9 * (9 - 1)) / 2 = (9 * 8) / 2 = 72 / 2 = 36. Which is the next term. {1, 9, 36}. and then continue this until you have row+1 amount of numbers, or in other words repeat it "row" amount of times, since you already started with 1.
In doing so you will get {1, 9, 36, 84, 126, 126, 84, 36, 9, 1}
Now to use binomial theorem to expand find the degrees. Cause currently we can determine the coefficients, but what about the degrees? So it's as if we have the equation: [tex]1(2x)^a(3)^b+9(2x)^c(3)^d+36(2x)^e(3)^f.....[/tex] and so on until we use up all the numbers in the row. But binomial theorem essentially states that the degrees will expand as such: [tex]C_0a^nb^0+C_1+a^{n-1}b^{0+1}+C_2a^{n-2}b^{0+2}...C_na^{n-n}b^{n}[/tex]. So essentially the degree of a starts at n (in this case 9), and and then from there continues to go down until it reaches 0, while the degree of b starts at 0 (so it's just 1) and continues to go up (by 1) until it reaches n
So expanding it out gives us:
[tex]1(2x)^9(3)^0 + 9(2x)^{9-1}(3)^{0+1}+36(2x)^{9-2}(3)^{0+2}+84(2x)^{9-3}(3)^{0+3}[/tex]... and so on But in this case we really only care about degree six which occurs at 9-3. So let's focus on that
[tex]84(2x)^6(3)^3[/tex]. This evaluates out to [tex]84(64x^6)(27)[/tex] which then evaluates to [tex]145,152x^6[/tex]
