How we can prove this vector identity?

Question

I was trying to rederive the formula of the angular momentum of electromagnetic field, and all the steps are clear for me except this one which I took from
"Photons and Atoms: Introduction to Quantum Electrodynamics" by Claude Cohen‐Tannoudji
$$sum_aUnabla_a(rtimesnabla)A_{bot a}=U(rtimesnabla)(nablacdot A_{bot})-U(nablatimes A_{bot})$$
Where $U$ is the Coulomb potential, $$A_{bot}$$ is the transvers vector potential and $ a$ is sub scripts run from 1 to 3. (like $x,y,z$).
Can any one help me how we get this result? I tried all vector identities I know.

ytlu · Accepted Answer

U and the perpendicular sub-index are irrelavent. I will omit them.
Some index expressions:
$$
nabla_a = frac{partial}{partial x_a} 
nabla cdot A = sum_i frac{partial A_i}{partial x_i}
   nabla times A = sum_k hat{e}_k sum_i sum_j epsilon_{k i j} frac{partial A_j}{partial x_i}
vec{r} = sum_i x_i hat{e}_i
   vec{r} times nabla = sum_k hat{e}_k sum_i sum_j epsilon_{k i j} x_i frac{partial}{partial x_j}
$$
where $hat{e}_k$ is the unit vector in $k$th direction,  and $ epsilon_{k i j}$ the Levi-Civita function. The Levi-Civita function is anti-symmetric, changes sign when exchange two indexes.
$$
sum_a nabla_a(rtimesnabla)A_{ a} 
= sum_a frac{partial}{partial x_a}(sum_k hat{e}_ksum_i sum_j epsilon_{a i j} x_ifrac{partial}{partial x_j})A_{ a} 
= sum_a sum_k hat{e}_k sum_i sum_j epsilon_{k i j} frac{partial x_i}{partial x_a} frac{partial}{partial x_j}A_{ a} 
 + sum_a sum_k hat{e}_ksum_i sum_j epsilon_{k i j} x_ifrac{partial}{partial x_a}frac{partial}{partial x_j}A_{ a} 
= sum_a sum_k hat{e}_k sum_i sum_j epsilon_{k i j} delta_{ai}frac{partial}{partial x_j}A_{ a} 
 + sum_k hat{e}_ksum_i sum_j epsilon_{k i j} x_ifrac{partial}{partial x_j} left{sum_a frac{partial A_a}{partial x_a} right} 
= sum_k hat{e}_k sum_i sum_j epsilon_{k i j} frac{partial}{partial x_j}A_{i} 
 + sum_k hat{e}_ksum_i sum_j epsilon_{k i j} x_ifrac{partial}{partial x_j} left{ sum_a frac{partial A_a}{partial x_a} right} 
= -sum_k hat{e}_k sum_j sum_i epsilon_{k j i} frac{partial A_i}{partial x_j} 
 + left{sum_k hat{e}_ksum_i sum_j epsilon_{k i j} x_ifrac{partial}{partial x_j} right} left{ sum_a frac{partial A_a}{partial x_a} right} 
= - (nablatimes A) + (rtimesnabla)(nablacdot A)
$$

How we can prove this vector identity?

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