Demostrating possible equivalence of two tensors

Is there anyway to see by inspection that a form like

$$a(x^2 )^{-3} (g _{μσ} x_{\rho} x_{ ν} + g_{μρ} x_{σ} x_{ ν} +g_{νσ} x_{ρ} x_{ μ} + g_{ νρ} x_{ σ} x_{ μ} )$$ may be equivalent to (i.e reduced down to or reexpressed) $$b(g _{μν} x_{ ρ} x_{ σ} + g_{ ρσ} x_{ μ} x_{ ν} )(x^2 )^{ −3} ?$$ where $g_{\mu \nu}$ is the metric tensor (diagonal).

I have tried to put in various permutations of $\mu \nu \rho \sigma$ and from $1111$ and $2222$ for example, I obtained the constraint that $a/b = 2$ but I am not really sure what this means. If I try the combination $1221$ e.g then it implies $b=0$, which seems to contradict my first result.

Does this mean that the two forms are not equivalent?

Answer

Indeed, if no values of $a$ and $b$ work for across different sets of indices, then the forms are not equivalent.

In fact, these two forms are not equivalent even under the restriction of the metric being diagonal (and thus are not equivalent under a general metric). The diagonal case is easy to analyze, and you gave a good set of indices to do it: $\mu\nu\rho\sigma = 1221$. Then the large parenthesized part of the first expression becomes $g_{11} x_2 x_2 + g_{22} x_1 x_1$. The middle terms drop out because they involve off-diagonal parts of the metric. However, both terms in the second expression also involve off-diagonal metric coefficients, so the second expression is identically $0$.

Given tensor components $T_{\mu\nu\rho\sigma}$ and $S_{\mu\nu\rho\sigma}$, we have $T_{1221} = g_{11} x_2 x_2 + g_{22} x_1 x_1$ while $S_{1221} = 0$, so clearly $T \not\propto S$.