Calculate $\mathbb EB_1B_2^2 B_3^3$ cleverly (expressing the result, e.g. in terms of $\mathcal N(0,1)$ moments), where $B_t$ is standard Brownian motion.
I know that in tasks like this, it's often useful to use $W_s = W_t + (W_s – W_t)$ (like for example here).
However, I don't know if my way is correct:
$$\begin{align}\mathbb E(B_1B_2^2B_3^3) &= \mathbb E(B_1B_3)\mathbb E(B_2^2)\mathbb E(B_3^2) \\ &= \mathbb E \left[ -\frac 12 (B_1-B_3)^2 + \frac 12 (B_1^2+B_3^2) \right] \cdot \mathcal N(0,2)\cdot \mathcal N(0,3) \\ &= \left[ -\frac 12 \mathbb E(B_1-B_3)^2 + \frac 12 \mathbb E (B_1^2)+\frac 12 \mathbb E(B_3^2) \right] \cdot \mathcal N(0,2)\cdot \mathcal N(0,3) \\ &=
\left[ – \frac 12 \mathcal N(0,3-1)+\frac 12 \mathcal N(0,1)+ \frac 12 \mathcal N(0,3) \right) \cdot \mathcal N(0,2)\cdot \mathcal N(0,3)\end{align}$$
Best Answer
Follow up the comment I wrote, I complete the method for computing your expectation. It suffices to expand the function of $X,Y, Z$ thanks to Wolfram Alpha $$\begin{align} X(X+Y)^2(X+Y+Z)^3 &= \color{red}{X^6} + 5 X^5 Y + 3 X^5 Z + \color{red}{10 X^4 Y^2} + 12 X^4 Y Z \\ &+ \color{red}{3 X^4 Z^2} + 10 X^3 Y^3 + 18 X^3 Y^2 Z + 9 X^3 Y Z^2 + X^3 Z^3 \\ &+ \color{red}{5 X^2 Y^4} + 12 X^2 Y^3 Z + \color{red}{9 X^2 Y^2 Z^2} + 2 X^2 Y Z^3 \\ &+ X Y^5+ 3 X Y^4 Z + 3 X Y^3 Z^2 + X Y^2 Z^3 \tag{1} \end{align}$$
From the independence of $X,Y,Z$: $$\mathbb{E}(X^i Y^j Z^k) = \mathbb{E}(X^i )\mathbb{E}(Z^k)\mathbb{E}(Z^k)$$ and the moment property of $\mathcal{N}(0,1)$: $$\mathbb{E}(T^p) = \cases{0 & p is odd\\ (p-1)!! & p is even} \hspace{1cm} \text{for } T\in \{X,Y,Z\}$$ you can deduce the result of $\mathbb{E}(X(X+Y)^2(X+Y+Z)^3)$: it suffices to keep the terms with even exponent (the red terms of $(1)$), then $$\mathbb{E}(X(X+Y)^2(X+Y+Z)^3) = 5!! + (10+3+5)\cdot 3!! + 9 = 5\cdot 3+18\cdot 3 + 9 = 78$$
Remark: Not sure this method is the most clever one, but in my opinion, it is the general method for this kind of problems.