Find `n` in the binomial `(2 3+1/(3 3))^n` , if the ration 7th term from the beginning to the 7 term from the end `1/6dot`

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Correct Answer - `n = 9`
`1/6 = (.^(n)C_(6)(2^(1//3))^(n-6)(3^(-1//3))^(6))/(.^(n)C_(n-6)(2^(1//3))^(n-6)(3^(-1//3))^(n-6))`
`6^(-1) = 6^(-4) xx 6^(n//3)= 6^(n//3-4) rArr n/3 - 4 = - 1 rArr n = 9`

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