I have the following Ito process: $X_t=e^{B_t-\frac{t}{2}}$ with $\{B_t\}_{t\geq 0}$ is a $\mathbb{P}$-Brownian motion and I want to find under which probability measure $\mathbb{Q}$ the process $X_t^3=e^{3B_t-\frac{3}{2}t}$ becomes a martingale. The differential form of $X_t^3$ is applying Ito: \begin{equation} dX_t^3=3X_t^3dt+3X_t^3dB_t=3X_t^3(dt+dB_t) \end{equation} I'm looking for $\mathbb{Q}$ such that $dB_t^{\mathbb{Q}}=dB_t^\mathbb{P}+dt$. Thus applying Girsanov theorem such $\mathbb{Q}$ does exist if: \begin{equation} \dfrac{d\mathbb{Q}}{d\mathbb{P}}\bigg\lvert_t=\exp\bigg({-\int_0^tdB_s^{\mathbb{P}}}-\frac{1}{2}\int_0^t ds\bigg) \end{equation} is a true martingale. This is easily verified since Novikov condition holds trivially because the process inside the expected value is purely deterministic. \begin{equation} E\bigg(\exp\bigg(\dfrac{1}{2}\int_0^tds\bigg)\bigg)<\infty \end{equation} Now I have that under $\mathbb{Q}$ the process $X_t^3$ is a $\mathbb{Q}$-Brownian motion.
Now I'm asked to compute the following: \begin{equation} E^{\mathbb{Q}}(X_t)=\int Xd\mathbb{Q}=\int X\dfrac{d\mathbb{Q}}{d\mathbb{P}}d\mathbb{P}=E^\mathbb{P}(X\dfrac{d\mathbb{Q}}{d\mathbb{P}}) \end{equation} So replacing the Radon Nykodym derivative with $e^{-B^\mathbb{P}_t-\frac{t}{2}}$ I got: \begin{equation} E^{\mathbb{Q}}(X_t)=E^\mathbb{P}(X\dfrac{d\mathbb{Q}}{d\mathbb{P}})=E^\mathbb{P}(e^{+B^\mathbb{P}_t-\frac{t}{2}}e^{-B^\mathbb{P}_t-\frac{t}{2}})=e^{-t}E(e^{B_t^\mathbb{P}-B_t^{\mathbb{P}}})=1 \end{equation} where I used that $B_t^\mathbb{P}-B_t^\mathbb{P}\sim N(0,2t)$.
Now my problem is that I strongly think that the last assumption is not correct because the two Brownian motion are the same random variable and actually the difference is zero isn't it?