The basic algebraic problem

Both $3s$ cancel each other and we are left with $17+17=\boxed{34}$

$Sol.\\ \frac { 17+17 }{ 3 } \times 3\\ "3"\quad Cancelled\quad with\quad "3"\\ and\quad 17+17=34$

$\frac{17+17}{3} \times 3=17+17=34$

The $3s$ canceled each other

THe denominators nicely cancel out to yield $17 + 17 = \color{#69047E}{\boxed{34}}$ .

Because you divided it, when you times it the outcome is just the starting equation of 17 + 17

Both the 3 cancel each other and the remaining value is the sum of 17+17=34

[(17+17)/3]*3=(17+17)=34

(17+17)/3) 3 => (34/3) 3 => 34