Let $B_t$ be a standard Brownian motion and let $M_t = B_t^2 -t$.
I want to show that $[M]=[B^2]$. Therefore I want to use the linearity of quadratic variations \begin{align} [\alpha X + \beta Y, Z] = \alpha[X,Z] + \beta [Y,Z]. \end{align} So, \begin{align} [M]_t &= [B^2 - t]_t \\ &= [B^2 - t, B^2 - t]_t \\ &= [B^2, B^2 - t]_t - [ t, B^2 - t]_t \\ &= [B^2, B^2]_t - [ t, B^2]_t - \big( [ B^2, t]_t - [t,t] \big) \\ &= [B^2]_t -2 [ B^2, t]_t + [t]. \end{align} How to find that $[M]_t = [B^2]_t$?