Show that $\mu_r = \mu_r' - \binom{r}{1} \mu'_{r-1} \cdot \mu + \cdots + (-1)^i \binom{r}{i} \mu'_{r-i} \cdot \mu^i + \cdots + (-1)^{r-1}(r-1)\cdot \mu^r$ for r = 1, 2, 3,,..., and use this formula to express $\mu_3$ in terms of moments.
Where $\mu_r = E[(X-\mu)^r]=\sum_x(x-\mu)^r \cdot f(x)$ which is the rth moment about the mean.
We also have $$\frac{d^rM_X(t)}{dt^r}|_{t=0}=\mu'_r$$
I've tried expanding the expected value as $(X-\mu)^r=\sum_{i=0}^{\infty} (-1)^{i+1} \binom{r}{i} \mu^rX^{r-i}$ so when $\mu_r$ I should get $$(X-\mu)^3 = \sum_{i=0}^{3}(-1)^{i+1} \binom{r}{i} \mu^rX^{r-i}$$ for r = 3 $$\implies E[(X-\mu)^3] = E\left[(1)\binom{3}{0} \mu^3 +(-1)\binom{3}{1}\mu^2 X+(1)\binom{3}{2} \mu X^2+(-1)\binom{3}{3} X^3 \right]$$
$$\implies E[(X-\mu)^3] = E(X) - 2\mu E(X^2) +3\mu^2E(X)^3$$
I'm unsure as to whether this was the formula the question was asking for?