If I have two correlated normal distributed random variables, i.e. $A \sim N(m_a,\sigma_a^2)$ and $B \sim N(m_b,\sigma_b^2)$, with correlation coefficient $ Corr(A,B) =\rho $, then I know that the $Cov(A,B) = \rho \sigma_a \sigma_b$.
My question: What ist the covariance of $A$ and $B^2$, i.e. $Cov(A,B^2)$ ? My approach: For simplicity, we can also assume that $B$ is standard normal, i.e. $m_b = 0$ and $\sigma^2_b = 1$. Then we have: \begin{align} Cov(A,B^2) &= E\left[(A-m_a)(B^2-E(B^2)) \right] \\ &= E\left[(A-m_a)(B^2-1) \right] \\ &= E[AB^2] - E[A] - m_aE[B^2] + m_a, \hspace{0.5cm} \text{since $E[B^2]=1$} \\ &= E[AB^2] - m_a - m_a + m_a \\ &= E[AB^2] - m_a \end{align} At this point I don't get further.