How does the variance appear in $\frac{E(X-\mu)^3}{\sigma^3}$

Question

I wanted to know how the variance appears in the following moment about the mean when calculating for skewness: $\frac{E(X-\mu)^3}{\sigma^3}$

By expanding it we get:

$$\frac{E(X^3)-3\mu[E(X^2)+\mu E(X)]-\mu^3}{\sigma^3} $$

I understand that $Var(X) = \sigma^2 = E[X^2]-\mu^2$

Though I do not understand why $E(X^2)+\mu E(X)=\sigma^2$ and would really appreciate some clarification.

Answer 1

Note that $E[X] = \mu$, hence $$ E[X^2] - \mu E[X] =E[X^2] - E[X] E[X] = E[X^2] - E^2[X] = Var(X) =\sigma^2 $$ Therefore, the skewness is \begin{align} \frac{E(X-\mu)^3}{\sigma^3} &= \frac{E[X^3] - 3\mu E[X^2] + 3\mu^2 E[X] - \mu^3}{\sigma^3}\\ & = \frac{E[X^3] - 3\mu ( E[X^2] - \mu E[X]) - \mu^3}{\sigma^3}\\ & = \frac{E[X^3] - 3\mu \sigma^2 - \mu^3}{\sigma^3} \end{align}