Edit:

Sorry, I didn't see the minus sign. You should use the following formula:

`(x-y)^n=sum_(k=0)^n(-1)^k((n),(k))x^(n-k)y^k`

So in the end the term independent of `x` is `(-1)^5((15),(5))3^10 2^5=-5674372704`

Once again I'm sorry for my mistake.

We will use binomial expansion formula:

`(x+y)^n=sum_(k=0)^n((n),(k))x^(n-k)y^k`

Therefore we have

`(3x-2/x^2)^15=sum_(k=0)^15((15),(k))(3x)^(15-k)(2/x^2)^k`

We need to find term for which the exponents of both `x` terms are equal so that they can cancel out i.e. we need to find `k` such that

`x^(15-k)=x^(2k)`

`15-k=2k`

`3k=15`

`k=5`

Thus we get our term for `k=5.`

`((15),(5))(3x)^(15-5)(2/x^2)^5=(15cdot14cdot13cdot12cdot11)/(1cdot2cdot3cdot4cdot5)3^10x^10cdot2^5/x^10=`

`3003cdot59049cdot32=5674372704`

**The term independent of `x` is `((15),(5))3^10 2^5=5674372704.`**

**Further Reading**