Let $\mathbf B=\{B_t\}_{t\ge0}$ a continuous Brownian motion, what is then $\large\mathbf E[B_s^4B_t^2-2B_sB_t^5+B^6_s]$, for $t\ge s$ ?
How can I factorize the expression in the parenthesis, If the last summand were $B_t^6$ instead of $B_s^6$ the expectation would be equal to $\mathbf E[(B_s^2B_t-B_s^3)^2]$ and $E[B_s^2B_t-B_s^3]=0$, can we derive something from there, maybe a typo ?
Since $B_t$ is Gaussian with mean $0$ and variance $t$, the moments of $B_t$ can be calculated explicitly. For any $k \in \mathbb{N}$ we have
$$\mathbb{E}(B_t^{2k+1}) = 0 \tag{1}$$
and
$$\mathbb{E}(B_t^{2k}) = t^k \frac{2^k \Gamma(k+1/2)}{\sqrt{\pi}},$$
in particular,
$$\mathbb{E}(B_t^2) = t \qquad \mathbb{E}(B_t^4) = 3t^2 \qquad \mathbb{E}(B_t^6) = 15 t^3. \tag{2}$$
Back to your exercise: We have
$$\begin{align*} B_s^4 B_t^2 - 2 B_s B_t^5 + B_t^6 &= B_t^2 (B_t^2-B_s^2)^2 = B_t^2 (B_t-B_s)^2 (B_t+B_s)^2 \\ &= ((B_t-B_s)+B_s)^2 (B_t-B_s)^2 (B_t+B_s)^2 \\ &= \left[ (B_t-B_s)^4 + 2 (B_t-B_s)^3 B_s + (B_t-B_s)^2 B_s^2 \right] ((B_t-B_s)+2B_s)^2 \end{align*}$$
Expanding also the square $((B_t-B_s)+2B_s)^2$ in the last line, we get a lot of terms. However, using that the odd moments equal $0$, see $(1)$, and using the independence and stationarity of the increments of Brownian motion, this boils down to
$$\begin{align*} \mathbb{E}( B_s^4 B_t^2 - 2 B_s B_t^5 + B_t^6 ) &= \mathbb{E}(B_{t-s}^6) + 4 \mathbb{E}(B_{t-s}^4) \mathbb{E}(B_s^2) + 8 \mathbb{E}(B_{t-s}^4) \mathbb{E}(B_s^2) \\ &\quad + \mathbb{E}(B_{t-s}^4) \mathbb{E}(B_s^2) + 4 \mathbb{E}(B_{t-s}^2) \mathbb{E}(B_s^4) \\ &\stackrel{(2)}{=} 15 (t-s)^3 + 39 (t-s)^2 s + 4 (t-s) s^2. \end{align*}$$
Finally, since
$$\mathbb{E}(B_t^6 -B_s^6) \stackrel{(2)}{=} 15 (t^3-s^3),$$
we get
$$ \mathbb{E}( B_s^4 B_t^2 - 2 B_s B_t^5 + B_s^6 ) = 15 (t-s)^3 + 39 (t-s)^2 s + 4 (t-s) s^2 - 15 (t^3-s^3)$$