The value `[(1)/(6)((2log_(10)(1728))/(1+(1)/(2)log_(10)(0.36)+(1)/(3)log_(10)8))^(1//2)]^(-1)` is :

Answer

To solve the given expression \(\left[\frac{1}{6}\left(\frac{2\log_{10}(1728)}{1+\frac{1}{2}\log_{10}(0.36)+\frac{1}{3}\log_{10}(8)}\right)^{\frac{1}{2}}\right]^{-1}\), we will follow these steps:

### Step 1: Simplify \(\log_{10}(1728)\)
First, we can express \(1728\) in terms of its prime factors:
\[ 1728 = 12^3 \]
Thus,
\[ \log_{10}(1728) = \log_{10}(12^3) = 3\log_{10}(12) \]

### Step 2: Simplify the denominator
Now we simplify the denominator:
\[ 1 + \frac{1}{2}\log_{10}(0.36) + \frac{1}{3}\log_{10}(8) \]
We can express \(0.36\) and \(8\) in terms of their logarithms:
\[ 0.36 = \left(\frac{6}{10}\right)^2 = \frac{36}{100} \Rightarrow \log_{10}(0.36) = \log_{10}(36) - \log_{10}(100) = 2\log_{10}(6) - 2 \]
\[ 8 = 2^3 \Rightarrow \log_{10}(8) = 3\log_{10}(2) \]
Substituting these values back, we get:
\[ 1 + \frac{1}{2}(2\log_{10}(6) - 2) + \frac{1}{3}(3\log_{10}(2)) = 1 + \log_{10}(6) - 1 + \log_{10}(2) = \log_{10}(6) + \log_{10}(2) \]
Using the property of logarithms:
\[ \log_{10}(6) + \log_{10}(2) = \log_{10}(12) \]

### Step 3: Substitute back into the expression
Now substituting back into our expression:
\[ \frac{2\log_{10}(1728)}{1+\frac{1}{2}\log_{10}(0.36)+\frac{1}{3}\log_{10}(8)} = \frac{2(3\log_{10}(12))}{\log_{10}(12)} = 6 \]

### Step 4: Final expression
Now we substitute this back into the original expression:
\[ \left[\frac{1}{6}(6)^{\frac{1}{2}}\right]^{-1} = \left[\frac{1}{6} \cdot \sqrt{6}\right]^{-1} = \frac{6}{\sqrt{6}} = \sqrt{6} \]

### Final Answer
Thus, the value of the given expression is:
\[ \sqrt{6} \]