Answer:
[tex]-540[/tex]
Step-by-step explanation:
From binomial theorem, we can know
first term would be x^6
2nd term x^5
3rd term x^4
4th term x^3
Also, from binomial theorem, we can know
first term would be y^0
2nd term would be y^1
3rd term would be y^2
4th term would be y^3
We can see that we are looking for the 4th term (x^3y^3).
To find coefficient of 4th term, we use write:
[tex]6C3(3x)^{6-3}(-y)^3[/tex]
We can expand 6C3 using formula: [tex]nCr= \frac{n!}{(n-r)!r!}[/tex]
Now, we have:
[tex]6C3(3x)^{6-3}(-y)^3\\(\frac{6!}{(6-3)!3!})(3x)^3(-y)^3\\(20(27x^3)(-y^3)\\-540x^3y^3[/tex]
Thus, the coefficient is -540
