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.

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