Kronecker delta in the harmonic lattice potential energy

This is a rather mathematical question regarding a derivation of the harmonic lattice energy $Phi^{ham}$ in a book about the quantum theory of solids. Specifically, I found the third equivalence in Eqn. (1) below is quite bizarre:

$$begin{aligned}
Phi^{mathrm{harm}} &=Phi_{0}+frac{1}{4} sum_{i kappa, j nu} sum_{alpha beta}left(u_{i kappa, alpha}-u_{j nu, alpha}right) phi_{alpha beta}left(boldsymbol{R}_{i}+boldsymbol{d}_{kappa}-boldsymbol{R}_{j}-boldsymbol{d}_{nu}right)left(u_{i kappa, beta}-u_{j nu, beta}right)
&=Phi_{0}+frac{1}{2} sum_{i j} sum_{kappa alpha, nu beta} phi_{alpha beta}left(boldsymbol{R}_{i}+boldsymbol{d}_{kappa}-boldsymbol{R}_{j}-boldsymbol{d}_{nu}right)left(u_{i kappa, alpha} u_{i kappa, beta}-u_{i kappa, alpha} u_{j nu, beta}right)
&=Phi_{0}+frac{1}{2} sum_{i j} sum_{kappa alpha, nu beta}left{left[sum_{j^{prime} nu^{prime}} phi_{alpha beta}left(boldsymbol{R}_{i}+boldsymbol{d}_{kappa}-boldsymbol{R}_{j^{prime}}-boldsymbol{d}_{nu^{prime}}right)right] delta_{i j} delta_{kappa nu}right.
&left.-phi_{alpha beta}left(R_{i}+d_{kappa}-R_{j}-d_{nu}right)right} u_{i kappa, alpha} u_{j nu, beta}
&=Phi_{0}+frac{1}{2} sum_{i j} sum_{kappa alpha, nu beta} u_{i kappa, alpha} D_{kappa nu, alpha beta}left(R_{i}-R_{j}right) u_{j nu, beta}
end{aligned}tag{1}$$

where

$$Phi_{0}=frac{1}{2} sum_{i kappa neq j nu} phileft(boldsymbol{R}_{i}+boldsymbol{d}_{kappa}-boldsymbol{R}_{j}-boldsymbol{d}_{nu}right)$$
is the cohesive energy with $boldsymbol{R}_i$ and $boldsymbol{d}_{kappa}$ being lattice coordinates and base atomic coordinates. $phi$ is the interaction potential between two atoms. $u_{ikappa,alpha}$ represent the atomic displacement at $alpha$ direction, and
$$phi_{alpha beta}left(boldsymbol{R}_{i}+boldsymbol{d}_{kappa}-boldsymbol{R}_{j}-boldsymbol{d}_{nu}right)=frac{partial^{2} phileft(boldsymbol{R}_{i}+boldsymbol{d}_{kappa}-boldsymbol{R}_{j}-boldsymbol{d}_{nu}right)}{partial R_{i alpha} partial R_{i beta}}.$$

What confuses me in (1) is the part in the square brackets. The intention of the author is to use the product $u_{ikappa,alpha}u_{jnu,beta}$ throughout the derivation. The author later uses the terms in the curly bracket to define a matrix,
$$begin{aligned}
D_{kappa nu, alpha beta}left(R_{i}-R_{j}right)=&left[sum_{j^{prime} nu^{prime}} phi_{alpha beta}left(R_{i}+d_{kappa}-R_{j^{prime}}-d_{nu^{prime}}right)right] delta_{i j} delta_{kappa nu}
&-phi_{alpha beta}left(R_{i}+d_{kappa}-R_{j}-d_{nu}right)
end{aligned}tag{2}$$

But why is
$$left[sum_{j^{prime} nu^{prime}} phi_{alpha beta}left(boldsymbol{R}_{i}+boldsymbol{d}_{kappa}-boldsymbol{R}_{j^{prime}}-boldsymbol{d}_{nu^{prime}}right)right] delta_{i j} delta_{kappa nu}u_{i kappa, alpha} u_{j nu, beta}=phi_{alpha beta}left(boldsymbol{R}_{i}+boldsymbol{d}_{kappa}-boldsymbol{R}_{j}-boldsymbol{d}_{nu}right)left.u_{i kappa, alpha} u_{i kappa, beta}right.tag{3}$$

I was trying to use the properties of Kronecker delta to understand (3) but couldn’t find any useful explanation so far.