Using the binomial theorem we can expand (x^2+1/x)^9

(x^2+1/x)^9

=>x^18 + 9x^16*(1/x) + 36x^14*(1/x^2) + 84x^12*(1/x^3) + 126x^10*(1/x^4) + 126x^8*(1/x^5) + **84x^6*(1/x^6)**+ 36x^4*(1/x^7) + 9*x^2*(1/x^8) + 1/x^9.

Therefore the term that does not have x is the one highlighted in the series above. Here only 84 is left as x in the first term which appears in the numerator is cancelled by the x in the second term which appears in the denominator.

The term with an x is the 7th term of the binomial expansion.

The binomial expansion of (A+B)^n = a^n+nC1*A^(n-1)*B+nC2*A^(n-2)*B2+....nCr-1*A^(n-r+1)*B^(r-1)+...+B^n.

So the rth term os Tr = nCr-1*A^(n-r+1)*B^(r-1).

In this case A = x^2, B = 1/x. n = 9.

If r is the term which does not have xt, then the exponent of x must be zero.

So Tr = 9Cr -1*(x^2)^(9-r+1)*x^(-1)(r-1)....(1)

So the exponent is 2(9-r+1)-r+1 = 0

18-2r+2-r+1 = 0

21- 3r = 0.

Therefore 21 = 3r, So r = 21/3 = 7.

Therefore T6 = 9C7 = (C2 = 9*8/2 = 36.

Therefore the 7th term in(x^2+x)^9 is 36.

To determine this term, we'll have to write the formula of the term that occupies the place k+1:

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

We'll identify the terms a and b from the development:

a = x^2 and b = 1/x

We don't know the place that the term occupies, so, we'll have to determine k.

n = 9

We'll substitute all we know in the formula (*);

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

We'll add the superscripts of x:

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

Since this term doesn't contain x, we'll impose the constraint that the superscript to be zero, because x^0 = 1.

[2(9-k)-k] = 0

We'll remove the brackets:

18 - 2k - k = 0

We'll combine like terms:

-3k = -18

We'll divide by -3:

**k = 6**

**The term that doesn't contain x occupies the 7th place, so it is the 7th term of development.**