Volumetric strain is three times longitudinal strain (cube)

Statement: for a cube under equal stress $\sigma$ in three mutually perpendicular directions (the same type of stress on all faces), the volumetric strain $\varepsilon_v = 3\varepsilon$, where $\varepsilon$ is the linear strain along any edge.

Proof. Let the cube have side $L$, so $V = L^3$. Under the stress, every edge changes by the same amount to $L + \delta L$, giving the linear strain

$$\varepsilon = \frac{\delta L}{L}$$

New volume:

$$V' = (L+\delta L)^3 = L^3\left(1+\varepsilon\right)^3 = V\left(1 + 3\varepsilon + 3\varepsilon^2 + \varepsilon^3\right)$$

Change in volume:

$$\delta V = V' - V = V\left(3\varepsilon + 3\varepsilon^2 + \varepsilon^3\right)$$

As the strains are very small, $\varepsilon^2$ and $\varepsilon^3$ are negligible compared with $\varepsilon$:

$$\varepsilon_v = \frac{\delta V}{V} \approx 3\varepsilon$$

Alternative (differentiation): from $V = L^3$, $dV = 3L^2\,dL$, so $\dfrac{dV}{V} = 3\dfrac{dL}{L}$, which gives the same result.

Note (with Poisson's effect): the strain along one edge caused by the three equal stresses is $\varepsilon = \dfrac{\sigma}{E} - \dfrac{\nu\sigma}{E} - \dfrac{\nu\sigma}{E} = \dfrac{\sigma}{E}(1-2\nu)$, so

$$\varepsilon_v = 3\varepsilon = \frac{3\sigma}{E}(1-2\nu) = \frac{\sigma}{K}$$

since $K = \dfrac{E}{3(1-2\nu)}$. Hence $\varepsilon_v = 3\varepsilon$ is proved.