Geometry

Hello, I just try see below i hope it's work,  My Fortiva

Let the side length of the inscribed cube be $s$. Since $\angle AOB = \angle AOC = \angle BOC = 90^\circ$, the triangle $ABC$ is a right triangle. Without loss of generality, assume $AB > AC$ and $BC > AC$. Let $D$ be the foot of the altitude from $O$ to $\triangle ABC$. Then $AD = BD = CD$ because $O$ is the circumcenter of $\triangle ABC$, and $OD = s\sqrt{3}/2$ because $O$ is the centroid of $\triangle ABD$.

[asy]
unitsize(1.5 cm);

pair A, B, C, D, E, F, O, P, Q, R;

A = (0,1);
B = (1,1);
C = (1,0);
O = (1/2,1/2);
D = (1/2,0);
E = (1/2,1);
F = (0,1/2);
P = (O + F)/2;
Q = (O + D)/2;
R = (O + E)/2;

draw(A--B--C--cycle);
draw(A--O);
draw(B--O);
draw(C--O);
draw(D--O);
draw(O--E,dashed);
draw(O--F,dashed);
draw(O--P,dashed);
draw(O--Q,dashed);
draw(O--R,dashed);
draw(E--F);
draw(P--Q);
draw(P--R);
draw(D--E,dashed);
draw(D--F,dashed);

label("$A$", A, N);
label("$B$", B, N);
label("$C$", C, E);
label("$D$", D, S);
label("$E$", E, N);
label("$F$", F, W);
label("$O$", O, N);
label("$P$", P, W);
label("$Q$", Q, S);
label("$R$", R, NE);
label("$s$", (O + R)/2, NE);
[/asy]

Let $E$ be the point on $\overline{AB}$ such that $BE = s$. Then $DE = BD - BE = AD - AE$, so
\begin{align*}
s^2 &= DE^2 \
&= (AD - AE)^2 \
&= AD^2 + AE^2 - 2 \cdot AD \cdot AE \
&= 3s^2/4 + (AB/2 - s)^2 - 2 \cdot 3s/4 \cdot (AB/2 - s) \
&= s^2/4 - 3sAB/8 + AB^2/4.
\end{align*}Solving for $s$, we find $s = AB/3$. It follows that $s = AC/3$ and $s = BC/3$ as well, so $s$ is the side length of the tetrahedron's other inscribed cube, and the desired side length is $s = AB/3 = AC/3 = BC/3 = \boxed{\frac{1}{3}}$.