Think logic 1

∛abc = 2, a+b+c = ?. abc = 8 (Cube both sides). Since a,b,c are integers, we can find the solution just finding the possible factors of 8 which are: 1,1,8 1,2,4 2,2,2 .Since a,b,c must be distinct (different), the only choice would be 1,2,4. Therefore 1+2+4 = 7

$\begin{aligned} \sqrt[3]{abc} & = 2\\ abc & = 8\\ & = 1×2×4\\ \implies a+b+c & = 1+2+4\\ & = \color{#3D99F6}{\boxed{7}} \end{aligned}$

$\quad Let\quad us\quad consider\quad three\quad no\quad a,b,c.\\ \quad All\quad three\quad numbers\quad are\quad positive\quad and\quad distinct\\ \quad so,\quad \\ \quad \quad By\quad A.M>G.M\\ \quad \frac { a+b+c }{ 3 } >\sqrt [ 3 ]{ abc } \\ \quad By\quad substituting\quad the\quad value\quad given\\ \quad \frac { a+b+c }{ 3 } >2\quad \\ \quad \Rightarrow a+b+c>6\\ Hence\quad answer\quad is\quad \boxed { 7 }$

$\sqrt [ 3 ]{ abc } =2\\ abc=8\\ a\cdot b\cdot c=1\cdot 4\cdot 2\\ a+b+c=1+4+2\\ a+b+c=7$

The integers a,b,c are distinct. So the set of numbers--1,2,4is the only possible answer.so 1+2+4=7.