ANSWER
2b
Step-by-step explanation
To simplify this expression, we have to apply some of the exponents' properties. First, the square root is a fractional exponent,
![√(x)=x^(1/2)](https://img.qammunity.org/2023/formulas/mathematics/college/ywa02tno250z2git3xw2hrsewiby80ugcw.png)
So we can rewrite the expression as,
![2√(a^2b^8)(ab^3)^(-1)=2(a^2b^8)^(1/2)(ab^3)^(-1)](https://img.qammunity.org/2023/formulas/mathematics/college/iv9pxnt0owz1r86tkuqx9qbttcqkxzjbgy.png)
Then, we can distribute the exponents into the multiplication,
![(xy)^z=x^zy^z](https://img.qammunity.org/2023/formulas/mathematics/college/su2awii3pi1vr6hoval3gp3roiozr7tshr.png)
In this problem,
![2(a^2b^8)^(1/2)(ab^3)^(-1)=2(a^2)^(1/2)(b^8)^(1/2)(a)^(-1)(b^3)^(-1)](https://img.qammunity.org/2023/formulas/mathematics/college/7kr21m5chxgndvtf1sun46w7wneas7xiyu.png)
Exponents of exponents are multiplied,
![(x^y)^z=x^(yz)](https://img.qammunity.org/2023/formulas/mathematics/high-school/9dnmtvm2ckikqf6ocyav1i132fidakeujl.png)
In this problem,
![2(a^2)^(1/2)(b^8)^(1/2)(a)^(-1)(b^3)^(-1)=2\cdot a^(2\cdot1/2)\operatorname{\cdot}b^{8\operatorname{\cdot}1/2}\operatorname{\cdot}a^(-1)\operatorname{\cdot}b^{3\operatorname{\cdot}(-1)}]()
Simplify if possible,
![2\cdot a^(2\cdot1/2)\operatorname{\cdot}b^{8\operatorname{\cdot}1/2}\operatorname{\cdot}a^(-1)\operatorname{\cdot}b^{3\operatorname{\cdot}(-1)}=2\cdot a^1\operatorname{\cdot}b^4\operatorname{\cdot}a^(-1)\operatorname{\cdot}b^(-3)]()
Now, the product of two powers with the same base is equal to the base raised to the sum of the exponents,
![x^y\cdot x^z=x^(y+z)](https://img.qammunity.org/2023/formulas/mathematics/college/6c0snt2363benbu4pfrypascyrw275qvay.png)
In this problem,
![2\cdot a^1\operatorname{\cdot}b^4\operatorname{\cdot}a^(-1)\operatorname{\cdot}b^(-3)=2\cdot a^(1-1)\operatorname{\cdot}b^(4-3)]()
Solve the subtractions,
![2\cdot a^(1-1)\operatorname{\cdot}b^(4-3)=2\cdot a^0\cdot b^1=2b]()
Hence, the simplified expression is 2b.