Answer:
9
Step-by-step explanation:
The first simplification we can make is to replace 6^0 with 1.
The first rules of exponents we can apply are ...
(a·b)^c = a^c·b^c . . . . . . similar to the distributive property
(a^b)^c = a^(b·c)
This reduces your expression to ...
[tex](2^8\cdot 3^{-5})^{-2}\cdot\left(\dfrac{3^{-2}}{2^3}\right)^4\cdot 2^{28}=2^{8(-2)}\cdot 3^{-5(-2)}\cdot\dfrac{3^{-2(4)}}{2^{3(4)}}\cdot 2^{28}\\\\=2^{-16}\cdot 3^{10}\cdot\dfrac{3^{-8}}{2^{12}}\cdot 2^{28}[/tex]
Now we can apply another two rules of exponents:
(a^b)(a^c) = a^(b+c)
1/a^b = a^-b
Using these, we have ...
[tex]=2^{-16-12+28}\cdot 3^{10-8}\\\\=2^0\cdot 3^2\\\\=1\cdot 9=9[/tex]
The value of the expression is 9.
