Let $A,B,C$ be $3\times 3$ matrices such that $A$ is symmetric and $B$ and $C$ are skew-symmetric.

Consider the statements

$\left(S1\right) A^{13} B^{26}-B^{26} A^{13}$ is symmetric

$\left(\mathrm{S}2\right) A^{26}{C}^{13}-C^{13} A^{26}$ is symmetric

Then,

Given,

Matrix $A$ is symmetric, $B$ is skew symmetric and $C$ is skew symmetric, 

So, $A^T=A, B^T=-B, C^T=-C$

Now let $M=A^{13}{B}^{26}-B^{26}{A}^{13}$

Then, $M^T={\left(A^{13} B^{26}-B^{26} A^{13}\right)}^T$

$={\left(A^{13}{B}^{26}\right)}^T-{\left(B^{26} A^{13}\right)}^T$

$={\left(B^T\right)}^{26}{\left( A^T\right)}^{13}-{\left(A^T\right)}^{13}{\left( B^T\right)}^{26}$

$=B^{26} A^{13}-A^{13} B^{26}=-M$

Hence, $M$ is skew symmetric

Now let, $N=A^{26}{C}^{13}-C^{13} A^{26}$

Then, $N^T={\left(A^{26}{C}^{13}\right)}^T-{\left(C^{13}{A}^{26}\right)}^T$

$=-{\left(C\right)}^{13}{\left(A\right)}^{26}+A^{26}{C}^{13}=N$

Hence, $N$ is symmetric.

So, only $S2$ is true.