$B$ is Brownian motion and $B_0 = 0$
Find the quadratic covariation between $[B^2,B^2]_t$
My attempt:
Using Ito's lemma on $B_t^2$ and $B_t^3$ we get:
$dB_t^2 = 2B_tdB_t + dt$
and
$dB_t^3 = 3B_t^2dB_t + 3B_tdt$
Thus $d[B^2,B^3]_t = (2B_tdB_t) (3B_t^2dB_t) + (2B_tdB_t) (3B_tdt) + dt (3B_t^2dB_t) + dt( 3B_tdt)$
using $dtdB_t = 0$ , $dB_tdB_t = dt$ and $dtdt = 0$
$d[B^2,B^3]_t = 6B_t^3dt$
hence $[B^2,B^3]_t = 6\int_0^t B_s^3 ds$
Am I right?