Think the other way

Let the largest of the three be $a$ . We have

$a^2+b^2+c^2=160 \\ a=b+c \\ b-c=4$

Plugging in $a=b+c$ in the first equation we get $2b^2+2bc+c^2=160 \longrightarrow b^2+bc+c^2=80$ , multiplying the result with $b-c=4$ we get $b^3-c^3=\boxed{320}$ .

The same method as Lakshya's Let\quad the\quad 3\quad positive\quad numbers\quad be\quad x,y,z\quad such\quad that\quad x<y<z\\ According\quad to\quad the\quad question,\\ \\ { x }^{ 2 }+{ y }^{ 2 }+{ z }^{ 2 }=160\\ x+y=z\\ y-x=4\\ Subsituting\quad the\quad value\quad of\quad z\quad in\quad the\quad first\quad equation,\quad we\quad have\quad { x }^{ 2 }+{ y }^{ 2 }+xy=80\\ Thus\quad { y }^{ 3 }-{ x }^{ 3 }=(y-x)({ x }^{ 2 }+{ y }^{ 2 }+xy)=4\times 80={ \boxed { 320 } }\\