If A and B are $n \times n$ matrices, define the Lie product $[A,B] = AB-BA$.
Exercise 1.37 of the book Basic Linear Algebra by T.S. Blyth and E.F. Robertson asks to prove that $$ (*) \ \ \ \ \ \ \ \ \ \ \ \ [[[A,B],C],D] + [[[B,C],D],A] + [[[C,D],A],B] + [[[D,A],B],C] = 0 $$ holds, but I couldn't prove it! Actually by expansion of the l.h.s of $(*)$ I reached to the point to show that $$[DB,AC] +[AC,BD] +[BD,CA] + [CA,DB] = 0$$ holds but I can't proceed any further.
Looks nice, I mean we can write like $$[D',A'] +[A',B'] +[B',C'] + [C',D'] = 0$$ and seeing 'neighboring' and 'cycling' but that doesn't become any simpler. Please help! Thank you.
Here are counterexamples for the identity $(*)$, with $$ A=\begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix},\; B=\begin{pmatrix} 1 & 0 \\ 1 & 0 \end{pmatrix},\; C=\begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix},\; D=\begin{pmatrix} d_1 & 1 \\ 0 & d_1 \end{pmatrix},\; $$ for all $d_1\in K$, where $K$ is a field of characteristic zero. We have $$ [[[A,B],C],D] + [[[B,C],D],A] + [[[C,D],A],B] + [[[D,A],B],C] =\begin{pmatrix} 0 & 2 \\ 0 & 0 \end{pmatrix}. $$ What is true instead is the Malcev-identity $$ [[A,B],[A,C]]=[[[A,B],C],A]+[[[B,C],A],A]+[[[C,A],A],B] $$