 Solution

In morphoelasticity, the deformation gradient factors as $\mathbf{F}=\mathbf{AG}$, where $\mathbf{G}$ accounts for local stress-free growth and $\mathbf{A}$ accounts for the subsequent elastic accommodation. Elastic incompressibility means that the latter preserves volume:It does not forbid volume increase by growth: here $\det \mathbf{F}=\det \mathbf{G}=\gamma$.

For the homogeneous diagonal deformation, the two unloaded transverse directions are equivalent. Taking the transversely symmetric branch implicit in the paper's hint gives ${\alpha }_1={\alpha }_2>0$. Set ${\alpha }_3=\alpha >0$. The determinant constraint is thensoEquality of the transverse stretches uses that symmetry assumption; elastic incompressibility alone would only give ${\alpha }_1{\alpha }_2{\alpha }_3=1$. The total deformation gradient and the actual axial length arebecause the initial cuboid is a unit cube.