Transforming joint Gaussians to independent random variables I saw this question in UIUC's probability class homework and it got me curious on how to solve it.

Suppose $X$ and $Y$ are jointly Gaussian such that $X \sim N(0, 9)$, $Y \sim N(0, 4)$; and the correlation coefficient is denoted by $\rho$. The solutions to the questions below may depend on $\rho$ and may fail to exist for some values of $\rho$.
For what values of $d$ is $X + dY$ independent of $(X - dY)^3$?

However I wasn't able to find a solution. How do you obtain the covariance of $(X - dY)^3$?  I am trying to do $Cov(Z-X,X^3)$ turns into $Cov(Z,X^3) - Cov(X,X^3)$ but I don't know what $Cov(X,X^3)$ solves to?
 A: There are two arguments you need to make.  First, if $A$ and $B$ are independent
random variables, then so are $g(A)$ and $h(B)$ independent random variables for
all (measurable) functions $g(\cdot)$ and $h(\cdot)$.  So, the question 
can be transformed into

For what values of $d$ are $X+dY$ and $X-dY$ independent random variables?

Second, since $X$ and $Y$ are jointly normal, so are $X+dY$ and $X-dY$ jointly
normal random variables, and so they are independent if $\text{cov}(X+dY,X-dY)$
equals $0$. Now,

$$\begin{align*}\text{cov}(X+dY,X-dY) &= \text{var}(X) -d^2\cdot\text{var}(Y) + d\cdot\text{cov}(Y,X)
-d\cdot\text{cov}(X,Y)\\
&= \text{var}(X) -d^2\cdot\text{var}(Y)
\end{align*}$$

since the covariance terms cancel, and so $\text{cov}(X+dY,X-dY) = 0$
for $d = \pm\sqrt{\frac{9}{4}}=\pm\frac{3}{2}$.
As a special case, note that if $\rho=\pm 1$, then 
$Y=\rho\frac{\sigma_Y}{\sigma_X}X = \pm\frac{2}{3}X$,
and in this case, one of the random variables $X+\frac{3}{2}Y$
and $X-\frac{3}{2}Y$ is zero with probability $1$.   
A: The answer is $d=\sqrt{\frac{9}{4}}$ and relies on a characterization of the normal distribution: 


*

*for two jointly normal random variable $X $and $Y$ with identical variance, $(X+Y)$ and $(X-Y)$ are independent normal random variables


user603 is specifically requested to not delete any of the boldfaced text in the above statement since deleting either part makes
the statement demonstrably incorrect.
A: You probably have to construct covariance matrix $\left(\begin{array}{cc}
9 & 6\rho  \\
6\rho & 4 \end{array}\right)$ and to find its eigenvalues and eigenvectors (both depending on $\rho$). Eigenvectors will led to the independent random variables expressed as linear combination of $X$ and $Y$:
$Z_1 = e_{11} X + e_{12} Y $ and $Z_2 = e_{21} X + e_{22} Y$.
