First thing to notice
In the third relation,
the powers on the right-hand side must be small enough to give a four-digit number in total. So there are only two options: either $T=2$, or $T=3$ and $S=1$. (Clearly $T$ isn't 0 or 1, as that would give too many leading zeros among the distinct digits.)
Option 1
If $T=2$, then in the third relation we must have $S\leq5$ otherwise the right-hand side would be more than four digits. So $S$ is one of 1,3,4,5.
The first relation is
$DE=S^2+4$. But $1^2+4=5$ (not two digits), $3^2+4=13$ (reusing digit 3), $4^2+4=20$ (reusing digit 2), $5^2+4=29$ (reusing digit 2). Contradiction!
Option 2
If $T=3$ and $S=1$, then the first relation is $1^3+3^3=28$, so $D=2$ and $E=8$.
The second relation is
$FGH=28+U^3+V^3$, where we cannot reuse the digits $1,2,3,8$. The only possibility is $28+5^3+7^3=496$, so we have $F=4,G=9,H=6$, and $\{U,V\}=\{5,7\}$ in some order.
Now the third relation is
$8128=496+9^3+11^3+13^3+1U^3$, which works with $U=5$.
Final solution
$S=1,D=2,T=3,F=4,U=5,H=6,V=7,E=8,G=9$.
$$28=1^3+3^3$$
$$496=1^3+3^3+5^3+7^3$$
$$8128=1^3+3^3+5^3+7^3+9^3+11^3+13^3+15^3$$