Suppose there is a cube of side of length ' $l$ '. When its temperature is increased by $\Delta \mathrm{T}$, it expands equally in all dimensions.

Hence from $\mathrm{V}=l^{3}$

$\Delta \mathrm{V}=(l+\Delta l)^{3}-l^{3}$

$=l^{3}+3 l^{2} \Delta l+3 l(\Delta l)^{2}+(\Delta l)^{3}-l^{3}$

But $(\Delta l)^{2}$ and $(\Delta l)^{3}$ are much more less than $l$, hence by neglecting them

$\Delta \mathrm{V}=3 l^{2} \Delta l$

… $(1)$

But, from linear expansion.

$\Delta l=\alpha_{l} l \Delta \mathrm{T}$

$\ldots$ $(2)$

$\therefore \quad$ By using value of equation $(1)$ in equation $(2)$,

$\Delta \mathrm{V} =3 l^{2}\left(\alpha_{l} l \Delta \mathrm{T}\right)$

$=3 l^{3} \alpha_{l} \Delta \mathrm{T}$

$\Delta \mathrm{V} =3 \mathrm{~V} \alpha_{l} \Delta \mathrm{T}$

$\left(\because l^{3}\right.=\text { volume of cube })$

$\ldots$ $(3)$

By comparing equation $(3)$ with general equation of volume expansion $\Delta \mathrm{V}=\alpha_{\mathrm{V}} \mathrm{V} \Delta \mathrm{T}$.

$\alpha_{V}=3 \alpha_{1}$ which is relation between coefficient of volume and linear expansion.