Show that $ S( a \times b) +(Sa) \times b \ + a \times (Sb) \ =0 $ Let $S$ be a $3 \times 3$ matrix which satisfies $S^T = S$ and $trace(S)=0$. Show for any $a,b \in \mathbb{R}^3$ that 
$$ S( a \times b) +(Sa) \times b \ + a \times (Sb) \ =0 $$
This can obviously done by a (long) brute force approach, but is there a more interesting way to approach this? Thank you!
 A: The problem becomes a little easier if we transform orthogonally all matrices and vectors to the basis where symmetric matrix $S$ becomes diagonal matrix $D$, vectors $a,b$ are still any vectors but expressed with reference to this basis, denote them $m,n$.  
In this basis
$$ D( m \times n) +(Dm) \times n \ + m \times (Dn) \ =0 $$
Vector product can be replaced with a use of skew-symmetric matrix, denote it $K$ .... $K(m)$ means here the skew-symmetric matrix assigned to the vector $m$ according to the rules in Wikipedia.
$$ D  K(m) n  +K(Dm) n \ + K(m) Dn  \ =0 $$
For any $n$
$$ (D  K(m)   +K(Dm)  \ + K(m) D)n  \ =0 $$
hence we need to prove that for any $m$
$$  D  K(m)   +K(Dm)  \ + K(m) D   =0 $$
Denote $D=\begin{bmatrix} d_1 & 0 & 0 \\
0 & d_2 & 0 \\
0 & 0 & d_3\\ \end{bmatrix}$ and $m=\begin{bmatrix} x & y & z \end{bmatrix}^T $
Then $Dm=\begin{bmatrix}  d_1x &  d_2y &  d_3z \end{bmatrix}^T $ and $K(m)= \begin{bmatrix} 0 & -z & y \\
z & 0 & -x \\
-y & x & 0\\ \end{bmatrix}$.
Consequently $DK(m)+K(m)D=\begin{bmatrix} 0 & -d_1z & d_1y \\
d_2z & 0  & -d_2x \\
-d_3y & d_3x & 0\\ \end{bmatrix}+\begin{bmatrix} 0 & -d_2z & d_3y \\
d_1z & 0 & -d_3x \\
-d_1y & d_2x & 0\\ \end{bmatrix}  = \begin{bmatrix} 0 & -(d_2+d_1)z & (d_3+d_1)y \\
(d_1+d_2)z & 0 & -(d_3+d_2)x \\
 -(d_1+d_3)y & (d_2+d_3)x & 0\\ \end{bmatrix} $
Under change of basis trace is preserved ( $\operatorname{trace}  (D) = \operatorname{trace}(S)) $  hence we have $d_1+d_2+ d_3=0$ what leads to   $$DK(m)+K(m)D= -K(Dm)$$ 
Therefore for any $m$    we have $$  D  K(m)   +K(Dm)  \ + K(m) D   =0 $$ .
A: This is the approach that I was saying.
The form
$$g(a,b)=S(a\times b)+Sa\times b+a\times Sb$$
is linear in $a$, linear in $b$, and anti-symmetric $g(a,b)=-g(b,a)$.
By the symmetry of $S$, and rotating those triple products.
$$\begin{align}
a\cdot g(a,b)&=a\cdot S(a\times b)+a\cdot(Sa\times b) + a\cdot(a\times Sb)\\
&=Sa\cdot(a\times b)+b\cdot(a\times Sa) + Sb\cdot(a\times a)\\
&=Sa\cdot(a\times b)+Sa\cdot(b\times a) + 0\\
&=Sa\cdot(a\times b)-Sa\cdot(a\times b)\\
&=0
\end{align}$$
Also $$b\cdot g(a,b)=-b\cdot g(b,a)=0$$

Intermission:
The multiplication $(a\times b)\cdot g(a,b)$ is essentially computing the trace of $S$ in the basis $a,b,a\times b$. Now, it seemed to me that it is even faster to just go to compute this part for the basis $e_1,e_2,e_3$, only since $S$ is already a matrix in this basis, which is moreover orthogonal.
Likewise, the computation above could just be done for $e_1,e_2,e_3$, which is all we need. But since the general case is as easy, I left it that way.

So we do, 
$$\begin{align}
e_3\cdot g(e_1,e_2)&=e_3\cdot Se_3+e_3\cdot (Se_1\times e_2)+e_3\cdot (e_1\times Se_2)\\
&=e_3\cdot Se_3+e_1\cdot Se_1+e_2\cdot Se_2\\
&=0\ \text{ because this is the trace of }S
\end{align}$$
From the two computations above $e_k\cdot g(e_i,e_j)$ is zero for all $i,j,k\in\{1,2,3\}$. 
Therefore, $g(e_i,e_j)=0$ for all $i,j\in\{1,2,3\}$, since it is orthogonal to all elements of the frame $e_1,e_2,e_3$.
Hence $g(a,b)=0$ for all $a,b$, since the bilinear form is zero at all elements of the basis $e_1,e_2,e_3$.
