Given a Black-Scholes Model and a derivative with payoff $S_{T}^{3}$ at time $T$. Check that the value of that derivative at time t is $V_{t} = g(t, T)S_{t}^{3}$, where $g(t, T)$ has to be determined.
I know the value of a derivative given its payoff can be written as $V_{t} = e^{-r(T-t)} E[X|F_{t}]$, where X is the payoff $S_{T}^{3}$, but I do not know how to follow from here.
Edit: First, define $V_{t} = F(t, S_{t})$, where $$F(t, S_{t}) = e^{-r(T-t)} E[(f(Se^{(T-t)(r-\sigma^{2}/2) + \sigma \sqrt{T - t} Z}$$ where $Z$ is a $N(0, 1)$. Then $V_{t} = e^{-r(T-t)} E[(Se^{(T-t)(r-\sigma^{2}/2) + \sigma \sqrt{T - t} Z})^{3}]$.
Under the risk-neutral measure in the Black-Scholes model, the stock satisfies the SDE: $$dS_t = r S_t dt + \sigma S_t dB_t$$ whose solution is $S_t = S_0 \exp \left( (r - \sigma^2 / 2)t + \sigma B_t \right)$. The value of this derivative instrument is: $$\begin{align*} V(t,x) &= e^{-r(T-t)} \mathbb{E}\left[ S_T^3 \, | \, S_t = x\right] \\ &= e^{-r(T-t)} \mathbb{E}\left[ x^3 \exp \left( 3(r - \sigma^2 / 2)(T-t) + 3\sigma (B_T - B_t) \right) \right] \\ &= e^{-r(T-t)} x^3 \exp \left( 3(r - \sigma^2 / 2)(T-t) \right) \mathbb{E}(\exp (3\sigma (B_T - B_t))) \\ &= e^{-r(T-t)} x^3 \exp \left( 3(r - \sigma^2 / 2)(T-t) \right) e^{\frac{9}{2} \sigma^2 (T-t)} \end{align*}$$ where in the last line we used the fact that for $X \sim N(0, \sigma^2)$, we have $E{e^{X}} = e^{\frac{\sigma^2}{2}}$. Thus, $V$ is of the required form.