I am trying to prove the following via the direct definition of Ito integrals (i.e. without the use of the Ito formula).
$$ \int^{t}_{0}B^{2}_{s}dB_{s}=\frac{1}{3}B^{3}_{t} - \int^{t}_{0}B_{s}ds$$
Progress so far
So Using the Definition of the Ito integral we can write $$\int^{t}_{0}B^{2}_{S}dB_{s}=\lim_\limits{\Delta t_{j} \ \to 0}\sum_\limits{j}B^{2}_{j}\Delta B_{j}$$ Where $\Delta B_{j}=[B_{j+1}-B_{j}]. $
From here i then started by trying to calculate $\Delta(B^{3}_{j})$ which i got
$$ \Delta(B^{3}_{j}) = B^{3}_{j+1}-B^{3}_{j} = (B_{j+1}-B_{j})^{3}+3B_{j}B_{j+1}\Delta B_{j}$$
However this doesn't really get the desired result i'm not sure how to proceed i assume i have calculated $\Delta (B^{3}_{j})$ incorrectly?