I was knowing this formula for vectors $a,b,c$:
$a \times (b \times c) = (a\cdot c)b - (a\cdot b)c$
But I was thinking of the differential operator as a vector and hence I thought the above formula would also work for
$\nabla \times (F \times G) $
but it did not rather we have the following:
$\nabla \times (F \times G)= (\nabla \cdot G)F - (F\cdot \nabla)G + (G\cdot \nabla)F - (\nabla\cdot F)G$
To see why this fails, examine the Wikipedia proof of the triple product expansion. Writing their work in more detail, we find: \begin{align} (u\times (v\times w))_x &= u_y(v_xw_y - v_yw_x) - u_z(v_zw_x - v_xw_z)\\ &=u_yv_xw_y - u_yv_yw_x - u_zv_zw_x +u_zv_xw_z \\ &=u_y\color{red}{v_x}w_y + u_z\color{red}{v_x}w_z - u_yv_y\color{magenta}{w_x}-u_zv_z\color{magenta}{w_x} \\ &= \color{red}{v_x}(u_yw_y + u_zw_z) - \color{magenta}{w_x}(u_yv_y + u_zv_z) \end{align} You can see that they took advantage of the commutativity of products of scalars. But operators generally do not commute with other things! As an easy example, $\frac{d}{dx}f(x)g(x) \neq f(x)\frac{d}{dx}g(x)$. You will notice that when $u$ is replaced with the operator $\nabla$, we will instead obtain: \begin{align} (\nabla\times (v\times w))_x &= \partial_y (v_xw_y - v_yw_x) - \partial_z(v_zw_x - v_xw_z)\\ &=\partial_y(v_xw_y)-\partial_y(v_yw_x) - \partial_z(v_zw_x) + \partial_z(v_xw_z)\\ &= (\partial_yv_x) w_y+v_x(\partial_y w_y) - (\partial_yv_y)w_x-v_y(\partial_yw_x)-(\partial_zv_z)w_x - v_z(\partial_zw_x)\\&\qquad\qquad+(\partial_zv_x)w_z + v_x(\partial_zw_z) \\ &= v_x(\partial_y w_y) + v_x(\partial_zw_z) - (\partial_yv_y)w_x - (\partial_zv_z)w_x + (\partial_yv_x) w_y + (\partial_zv_x)w_z\\&\qquad\qquad -v_y(\partial_yw_x)-v_z(\partial_zw_x) \\ &=(\partial_yw_y+\partial_zw_z)v_x - (\partial_yv_y+\partial_zv_z)w_x+(w_y\partial_y + w_z\partial_z)v_x - (v_y\partial_y +v_z\partial_z)w_x \\ &\vdots\\ &=(\nabla\cdot w)v_x - (\nabla\cdot v)w_x+(w\cdot\nabla)v_x - (v\cdot\nabla)w_x \end{align}
and similarly for the other components.