# Use the following expansion to find the following (x- 3/x^2)^9 containing a term x^6.

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We'll use the formula of general term of binomial expansion (a+b)^n:

T(k+1) = C(n,k)*[a^(n-k)]*b^k

Let a = x, b = -3/x^2, n = 9

T(k+1) = C(9,k)*[x^(9-k)]*[(-3/x^2)^k]

This term must contain x^6, therefore we'll consider just the power of x.

We'll use the negative power property:

(1/x^2)^k = x^(-2k)

x^(9-k)*x^(-2k) = x^6

We'll add the exponents from the left side:

x^(9-k-2k) = x^6

Since the bases both sides are matching, we'll equate the exponents:

9-3k = 6

We'll divide by 3:

3 - k = 2

We'll keep k to the left side and we'll subtract 3 both sides:

-k = -3 + 2

-k = -1

k = 1

**Since k = 1, therefore the second term of the expansion, T2, will contain x^6.**