Answer:
9
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 ...
![(2^8\cdot 3^(-5))^(-2)\cdot\left((3^(-2))/(2^3)\right)^4\cdot 2^(28)=2^(8(-2))\cdot 3^(-5(-2))\cdot(3^(-2(4)))/(2^(3(4)))\cdot 2^(28)\\\\=2^(-16)\cdot 3^(10)\cdot(3^(-8))/(2^(12))\cdot 2^(28)](https://img.qammunity.org/2020/formulas/mathematics/middle-school/r8b6jwqxawuk1f4aklbf1i7gx9oxleevtv.png)
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 ...
![=2^(-16-12+28)\cdot 3^(10-8)\\\\=2^0\cdot 3^2\\\\=1\cdot 9=9](https://img.qammunity.org/2020/formulas/mathematics/middle-school/w7y2mjj6qygv9ich8d4qp5lo1jbr4sl9lj.png)
The value of the expression is 9.