Ever so equable I

We want to find positive real numbers $x,y,z$ such that $xyz = 2(xy + xz + yz) = 4(x+y+z)$ , so that $x+y+z \; =\; u \hspace{2cm} xy + xz + yz \; = \; 2u \hspace{2cm} xyz \; = \; 4u$ for some real $u$ ., This means that the cubic polynomial $f_u(X) \; = \; X^3 - uX^2 + 2uX - 4u$ must have three positive real zeros $x,y,z$ . Thus we deduce that the cubic discriminant $\Delta_3 = -12u^4 + 112u^3 - 432u^2 \; = \; -4u^2(3u^2 - 28u + 108)$ must be nonnegative. SInce the roots are all positive, we must have $u > 0$ (so that the coefficients of $f_u(X)$ alternate in sign). But the quadratic $3u^2 - 28u + 108$ has discriminant $-512 < 0$ and hence we deduce that $3u^2 - 28u + 108 \,>\, 0$ for all $u$ . Thus the only value of $u$ for which $\Delta \ge 0$ is $u=0$ , which implies that $x=y=z=0$ , and we do not have a proper cuboid in this case.

Thus there are no cuboids with the required property.

Let $a$ , $b$ , $c$ be the dimensions of the cuboid. Then the volume $V$ , the surface area $S$ and the total edge length $L$ are:
$\begin{aligned} (1) \ \ \ \ \ V& =abc \\ (2) \ \ \ \ \ S& =2\left( ab+bc+ca \right) \\ (3) \ \ \ \ \ L& =4\left( a+b+c \right) \\ \end{aligned}$

We want $V=S=L$ .

Then,

$\begin{aligned} \left. \begin{matrix} \left( 1 \right),\left( 2 \right)\Rightarrow \dfrac{1}{c}=\dfrac{1}{2}-\dfrac{1}{a}-\dfrac{1}{b} \\ \ \ \\ \left( 1 \right),\left( 3 \right)\Rightarrow \dfrac{1}{c}=\dfrac{ab-4}{4a+4b} \\ \end{matrix} \right\} & \Rightarrow \dfrac{1}{2}-\dfrac{1}{a}-\dfrac{1}{b}=\dfrac{ab-4}{4a+4b} \\ & \ \ \\& \Leftrightarrow \left( {{b}^{2}}-2b+4 \right){{a}^{2}}+2b\left( 2-b \right)a+4{{b}^{2}}=0 \\ \end{aligned}$

This quadratic, in terms of $a$ , must have non-negative discriminant $Δ$ , but $\Delta =4{{b}^{2}}\left( -3{{b}^{2}}+4b-12 \right)<0$ , for all $b>0$ .

Hence, there are no triples of positive real numbers simultaneously satisfying $(1)$ , $(2)$ and $(3)$ , i.e. there are no cuboids with the required property. The answer is $\boxed{None}$ .