Difficulty in proof of relationship between $Y$, $B$, $\sigma$

On internet, I was looking for the proof regarding the following relationship between $$Y$$ young's modulus,$$\sigma$$ poisson's ratio, and $$B$$ bulk modulus: $$Y=3B(1-2\sigma)$$, and I came across the following proof, which I have difficulty in understanding it.

Consider a cube subjected to three equal stresses S on the three faces The total strain in one direction or along one face due to the application of volumetric stress S is given as $$\text{linear Strain} = S/Y – \sigma S/Y –\sigma S/Y = S/Y( 1-2\sigma)$$ So $$\text{linear Strain} =(S/E)( 1-2r)$$. Now $$\text{Volumetric Strain} = 3 \cdot \text{Linear Strain}$$. So $$\text{Volumetric Strain} = 3 (S/Y)( 1-2\sigma)$$ Now by definition of bulk modulus of elasticity K. We have $$K= \frac{\text{volumetric Stress}} {\text{volumetric Strain}}$$. $$K= \frac{S}{{(3(S/Y)( 1-2\sigma)}}$$ , So $$Y=3K(1-2\sigma)$$

I am having main doubt regarding linear strain equation, especially why $$–\sigma S/Y$$ is being added two times. Rest of the proof is understandable to me. So please help me there. Moreover I want to ask is there, any simpler proof for this equation? Also, while finding/researching, I can't come up with proof for another similar equation: $$Y=2\eta(1+\sigma)$$, where $$\eta$$ is shear modulus. So I want help here too.

Hre is a general set of relations:

Start from the isotropic stress-strain relation $$\sigma_{ij} = \lambda \delta_{ij} e_{kk} + 2\mu e_{ij}$$ in terms of the Lame constants $$\lambda$$, $$\mu$$. By considering particular deformations, we can express the more directly measurable bulk modulus, shear modulus, Young's modulus and Poisson's ratio in terms of $$\lambda$$ and $$\mu$$.

The bulk modulus bulk modulus $$\kappa$$ is defined by $$dP =-\kappa \frac {dV}{V},$$ where an infinitesimal isotropic external pressure $$dP$$ causes a change $$V\to V+dV$$ in the volume of the material. This applied pressure corresponds to a surface stress of
$$\sigma_{ij}=-\delta_{ij}\,dP$$. An isotropic expansion away from the origin displaces points $$x_i\to x_i+\eta_i$$ in the material so that $$\eta_i =\frac 13 \frac {dV}{V}x_i.$$ The strains $$e_{ij}= \frac 12 (\partial_i\eta_j+\partial_j \eta_i)$$ are therefore given by $$e_{ij} = \frac 13 \delta_{ij} \frac {dV}{V}.$$ Inserting this strain into the stress-strain relation gives $$\sigma_{ij}= \delta_{ij}(\lambda +\frac 23 \mu)\frac {dV}{V}= - \delta_{ij} dP.$$ Thus $$\kappa= \lambda+\frac 23 \mu.$$

To define the shear modulus, we assume a deformation $$\eta_1 = \theta x_2$$, so $$e_{12}=e_{21}=\theta/2$$, with all other $$e_{ij}$$ vanishing.

The applied shear stress is $$\sigma_{12}=\sigma_{21}$$. The shear modulus, is defined to be $$\sigma_{12}/\theta$$. Inserting the strain components into the stress-strain relation gives $$\sigma_{12} =\mu\theta,$$ and so the shear modulus is equal to the Lame constant $$\mu$$. We can therefore write the generalized Hooke's law as $$\sigma_{ij} = 2\mu(e_{ij} -{\textstyle{\frac 13}} \delta_{ij} e_{kk}) +\kappa e_{kk} \delta_{ij},$$ which reveals that the shear modulus is associated with the traceless part of the strain tensor, and the bulk modulus with the trace.

Young's modulus $$Y$$ is measured by stretching a wire of initial length $$L$$ and square cross section of side $$W$$ under a tension $$T=\sigma_{33}W^2$$.

We define $$Y$$ so that
$$\sigma_{33} = Y\frac {dL}{L}.$$ At the same time as the wire stretches, its width changes $$W\to W+dW$$. Poisson's ratio $$\sigma$$ is defined by $$\frac{dW}{W}=-\sigma \frac{dL}{L},$$ so that $$\sigma$$ is positive if the wire gets thinner as it gets longer. The displacements are $$\eta_3 = z \left(\frac {dL}{L}\right),\nonumber\\ \eta_{1}=x \left(\frac{dW}{W}\right) = - \sigma x \left(\frac{dL}{L}\right),\nonumber\\ \eta_{2} = y \left(\frac{dW}{W}\right)= - \sigma y \left(\frac{dL}{L}\right),$$ so the strain components are $$e_{33} = \frac {dL}{L},\quad e_{11}=e_{22} = \frac{dW}{W}=-\sigma e_{33}.$$ We therefore have $$\sigma_{33} = (\lambda(1-2\sigma) +2\mu)\left(\frac {dL}{L}\right),$$ leading to $$Y= \lambda(1-2\sigma)+2\mu.$$ Now, the side of the wire is a free surface with no forces acting on it, so $$0=\sigma_{22}=\sigma_{11} = (\lambda(1-2\sigma) -2\sigma \mu)\left(\frac {dL}{L}\right).$$ This tells us that $$\sigma =\frac 12 \frac{\lambda}{\lambda+\mu},$$ and $$Y= \mu\left(\frac{3\lambda+2\mu}{\lambda+\mu}\right).$$ The desiired relations $$Y= 3\kappa(1-2\sigma),\nonumber\\ = 2\mu(1+\sigma),$$ now follow by simple algebra from those above.