How do I get the following bias when using $\bar{y}^{3}$ as an estimator for $\mu^{3}$ where $y$ is iid with mean $\mu$ and variance $\sigma^{2}$: $$ \frac{3 \mu \sigma^{2}}{N}+\frac{\mathbb{E}\left[(y-\mu)^{3}\right]}{N^{2}} $$
I tried the following:
Given that $$ (\bar{y}- \mu)^{3} = \bar{y}^{3} - 3\bar{y}^{2}\mu + 3\bar{y}\mu^{2} - \mu^{3}\\ $$ I used this to open the following expectation: $$ E[\bar{y}^{3}- \mu^{3}] = E[(\bar{y}-\mu)^{3}-3\bar{y}^{2}\mu+3\bar{y}\mu^{2}] = $$ $$ E[(\bar{y}-\mu)^{3}] + 3\mu E[\bar{y}^{2} - \bar{y}\mu] $$ $$ E[(\bar{y}-\mu)^{3}] + 3\mu [E[\bar{y}^{2}] - \mu E[\bar{y}]] $$ However, I don't know how I can continue from here once $\bar{y} = \frac{1}{N} \sum y_{i}$ I tried to open this definition but with little success.
Note that $E[\bar{y}] = \mu$ and $V[\bar{y}] = E[\bar{y}^2] - \mu^2 = \sigma^2/n$. Further $$ (\bar{y} - \mu)^3= \bar{y}^3 - \mu^3 + 3 \mu^2 \bar{y} - 3 \mu \bar{y}^2 $$ and so from this we can see that $$ \bar{y}^3 - \mu^3 = (\bar{y} - \mu)^3 - 3 \mu^2\bar{y} + 3 \mu \bar{y}^2 $$ and therefore \begin{align*} E[\bar{y}^3] - \mu^3 &= E[(\bar{y} - \mu)^3] - 3 \mu^2 E[\bar{y}] + 3 \mu E[\bar{y}^2] \\ &= E[(\bar{y} - \mu)^3] - 3 \mu^3 + 3 \mu (\sigma^2/n + \mu^2) \\ &= E[(\bar{y} - \mu)^3] + 3 \sigma^2\mu/n \end{align*}
Now, $$ E[(\bar{y} - \mu)^3]= E \left [ \left(\frac{1}{n}\sum_i (y_i -\mu)\right)^3 \right ]= \frac{1}{n^3} \left[\sum_i E(y_i -\mu)^3\ \right ] =\frac{1}{n^2} E(y -\mu)^3 $$ where the last two equalities hold by independence and identical distribution, respectively.