Proving $$\text{div}(\mathbf{F} \times \mathbf{G}) = \mathbf{G} \cdot \text{curl}(\mathbf{F}) - \mathbf{F}\cdot \text{curl}(\mathbf{G}).$$ I calculated the left hand side but its not the same as the right hand side. The right hand side has the coefficient of 2 multiplied by each term
Here's the problem:
A more advanced approach, just for the sake of interest and future needs.
$$ \nabla \cdot (F\times G) = \sum_{i,j,k} \frac{\partial}{\partial x_i} (\epsilon_{ijk} u_j v_k) = \sum_{i,j,k} \epsilon_{ijk}\left(G_k\frac{\partial F_j}{\partial x_i} + F_j\frac{\partial G_k}{\partial x_i} \right), $$
whereas $$ (\nabla \times F)\cdot G - (\nabla\times G)\cdot F = \sum_{i,j,k}G_i\epsilon_{ijk} \frac{\partial F_k}{\partial x_j} - \sum_{i,j,k}F_i\epsilon_{ijk} \frac{\partial G_k}{\partial x_j}. $$
Now observe:
$$ \sum_{i,j,k}G_i\epsilon_{ijk} \frac{\partial F_k}{\partial x_j} = - \sum_{i,j,k}G_i\epsilon_{jik} \frac{\partial F_k}{\partial x_j} = (-1)^2 \sum_{i,j,k}G_i\epsilon_{jki} \frac{\partial F_k}{\partial x_j} \equiv \sum_{i,j,k}G_k\epsilon_{ijk} \frac{\partial F_j}{\partial x_i}, $$
where I have just reindexed in the last equality. Similarly:
$$ - \sum_{i,j,k}F_i\epsilon_{ijk} \frac{\partial G_k}{\partial x_j} = (-1)^2 \sum_{i,j,k}F_i\epsilon_{jik} \frac{\partial G_k}{\partial x_j} \equiv \sum_{i,j,k}F_j\epsilon_{ijk} \frac{\partial G_k}{\partial x_i}. $$
Hence:
$$ (\nabla \times F)\cdot G - (\nabla\times G)\cdot F = \sum_{i,j,k}F_j\epsilon_{ijk} \frac{\partial G_k}{\partial x_i} + \sum_{i,j,k}G_k\epsilon_{ijk} \frac{\partial F_j}{\partial x_i} = \sum_{i,j,k}\epsilon_{ijk}\left( G_k\frac{\partial F_j}{\partial x_i} + F_j\frac{\partial G_k}{\partial x_i}\right) =$$ $$ \nabla \cdot (F\times G), $$
such as we want.
Remember: in the first equation I have used the Leibniz rule to achive two terms. The second trick is a game with indices and properties of the Levi-Civita symbol $\epsilon_{ijk}$.