Suppose I have $X_1,...X_{k} $ be random variables from the standard normal distribution. And I want to find the quantity $E[\overline X \overline X W\overline X + \overline X \overline XW^2]$, where $\overline X$ is the sample mean and $W$ is the sample variance, $W=S^2$. My attempt is to use linearity to make it becomes $E[\overline X \overline X W\overline X] + E[\overline X \overline XW^2]$. We know sample mean and sample variance is independent, so = $E[\overline X^3]E[W] + E[\overline X \overline X]E[W^2] $. Here, I know the expectation of sample mean squared(3rd term) is just the mean of a chi-square distribution with $r=1$ and the second term is just $\sigma^2$.
However, I am struggling to find $E[\overline X^3]$ because to my knowledge there is no distribution that is equal to $\overline X^3$ , same reason for $E[W^2]$. Is there a easy way to compute them? or do I have to actually take the derivative of the mgf from the normal distribution of $\overline X$ three times? Since $(\overline X=N(\mu,\sigma^2/k))$
$\bar X$ is a normal random variable $\mathcal N(\mu,\sigma^2/k)$, it's third moment is $\mu^3+\frac{3}{k} \mu \sigma^2$ (See Moments of normal random variable).
For $W$, observe that $U=W/\sigma^2$ is a Chi-squared of degree $k$, we know it's mean is $k$ and variance is $2k$, such that $\mathbb E[U^2]=2k+k^2$ and $\mathbb E[W]=\sigma^4(2k+k^2)$.