Let's take the first number they gave you: 9^867
If you think about it, this can really be written as (3^2)^867, because the 3^2 would result in 9, and then you would have 9^867.
Let's do this to the second number they gave you too: 9^428 = (3^2)^428
Now we have:
(3^2)^867 x (3^2)^428 x 3^-358 x 3^-2228
Now, we have to remember this rule:
Note that b and c could be any numbers, this just shows you the rule.
This means that (3^2)^867 is really 3^(2*867), which is 3^1734.
Let's do this to the second number too. ⇒ (3^2)^428 = 3^(2*428) = 3^856.
And now we have:
3^1734 x 3^856 x 3^-358 x 3^-2228
Like you said, we can now add up these exponets.
$$\begin{array}{lll}
3^{1734} \times 3^{856} \times 3^{-358} \times 3^{-2228}\\
3^{(1734)+(856)+(-358)+(-2228)}\\
3^{(2590)+(-2586)}\\
3^{2590-2586}\\
3^4\\
3 \times 3 \times 3 \times 3\\
9 \times 9\\
81
\end{array}$$
Ah, you beat me to it CPhill.
Well, looks like we got the same answer, so that's good. :)
