Find the value of
`(3^(-1) + 4^(-1) + 5^(-1) )^(0)`

Answer

To solve the expression \((3^{-1} + 4^{-1} + 5^{-1})^{0}\), we can follow these steps:

### Step 1: Understand the exponent
Any non-zero number raised to the power of zero is equal to 1. Therefore, we need to check if the expression inside the parentheses is non-zero.

### Step 2: Calculate the individual terms
We start by calculating each term in the expression:
- \(3^{-1} = \frac{1}{3}\)
- \(4^{-1} = \frac{1}{4}\)
- \(5^{-1} = \frac{1}{5}\)

### Step 3: Combine the fractions
Now we need to add these fractions together:
\[ 3^{-1} + 4^{-1} + 5^{-1} = \frac{1}{3} + \frac{1}{4} + \frac{1}{5} \]

To add these fractions, we need a common denominator. The least common multiple (LCM) of 3, 4, and 5 is 60.

### Step 4: Convert each fraction to have the common denominator
Now we convert each fraction:
- \(\frac{1}{3} = \frac{20}{60}\) (since \(1 \times 20 = 20\) and \(3 \times 20 = 60\))
- \(\frac{1}{4} = \frac{15}{60}\) (since \(1 \times 15 = 15\) and \(4 \times 15 = 60\))
- \(\frac{1}{5} = \frac{12}{60}\) (since \(1 \times 12 = 12\) and \(5 \times 12 = 60\))

### Step 5: Add the fractions
Now we can add these fractions:
\[ \frac{20}{60} + \frac{15}{60} + \frac{12}{60} = \frac{20 + 15 + 12}{60} = \frac{47}{60} \]

### Step 6: Substitute back into the original expression
Now we substitute this result back into the original expression:
\[ (3^{-1} + 4^{-1} + 5^{-1})^{0} = \left(\frac{47}{60}\right)^{0} \]

### Step 7: Apply the exponent rule
Since any non-zero number raised to the power of zero is equal to 1, we have:
\[ \left(\frac{47}{60}\right)^{0} = 1 \]

### Final Answer
Thus, the value of \((3^{-1} + 4^{-1} + 5^{-1})^{0}\) is **1**.

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