Cubic expectation of peanuts on squares

There are 9 squares on the table, and we drop 9 peanuts to the table one at a time. Each peanut fall into every square with equal probability 1/9. After dropping all peanuts, we denote $x_{i}$ as the number of peanuts on the i-th square, i=1,2,3...9, and define Y = $x_{1}^{3}+x_{2}^{3}+...+x_{9}^{3}$. Calculate expectation of Y.

Updates:

Based on linearity of expectation, permutation symmetry of $x_{i}$ and $x_{i}$ obeys Binomial distribution

$$E[Y]=\sum_{i=1}^{9}{E[X_{i}^3]}=\sum_{i=1}^{9}{n(n-1)(n-2)p^3+3n(n-1)p^2+np}$$

Given n=9 and p=1/9 for binomial distribution, we can derive:

$$E[Y]=9*(4+29/81)=39\frac{2}{9}$$

This answer should lie between $Y_{min}=9$ and $Y_{max}=9^3=729$.

For third order moment of binomial distribution, refer to The 3rd raw moment of a binomial distribution

Based on linearity of expectation, permutation symmetry of $x_{i}$ and $x_{i}$ obeys Binomial distribution

$$E[Y]=\sum_{i=1}^{9}{E[X_{i}^3]}=\sum_{i=1}^{9}{n(n-1)(n-2)p^3+3n(n-1)p^2+np}$$

Given n=9 and p=1/9 for binomial distribution, we can derive:

$$E[Y]=9*(4+29/81)=39\frac{2}{9}$$

This answer satisfy $Y_{min}=9<E[Y]<Y_{max}=9^3=729$.

For third order moment of binomial distribution, refer to The 3rd raw moment of a binomial distribution