Math question

4*3^-1 → 4/3

6*3^-2 → 6/9 → 2/3

Hence:

2 - 4*3^-1 - 6*3^-2 → 2 - 4/3 - 2/3  → 2 - 6/3  → 2 - 2  → 0

.

2-4*3^-1-6*3^-2 this is = 0 but why? can someone explain?

\( 2-4\times \frac{1}{3} -6\times \frac{1}{3^{2} } \)

= \(2-\frac{4}{3} -\frac{6}{9}\)

= \(2-\frac{18}{9} \)

= \(2-2\)

= \(0\)

\( More \ does \ not \ remain. \) 

asinus :- )  !