Question

Find the coefficient of the x^3y^3 term in the expansion of (3x-y)^6

Answer 1

`-540`

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:

`6C3(3x)^(6-3)(-y)^3`

We can expand 6C3 using formula:
`nCr= (n!)/((n-r)!r!)`

Now, we have:

`6C3(3x)^(6-3)(-y)^3\\((6!)/((6-3)!3!))(3x)^3(-y)^3\\(20(27x^3)(-y^3)\\-540x^3y^3`

Thus, the coefficient is -540