If 2160 = `2^(a). 3^(b) . 5^(c )`, find a, b and c. Hence calculate the value of `3^(a) xx 2^(-b) xx 5^(-c )`.

Answer

To solve the problem, we need to factorize the number 2160 into its prime factors and then express it in the form \(2^a \cdot 3^b \cdot 5^c\). After finding the values of \(a\), \(b\), and \(c\), we will calculate the expression \(3^a \cdot 2^{-b} \cdot 5^{-c}\).

### Step-by-Step Solution:

1. **Factorization of 2160**:
- Start dividing 2160 by 2 (the smallest prime number):
- \(2160 \div 2 = 1080\)
- \(1080 \div 2 = 540\)
- \(540 \div 2 = 270\)
- \(270 \div 2 = 135\) (135 is not divisible by 2 anymore)
- Now, divide by 3:
- \(135 \div 3 = 45\)
- \(45 \div 3 = 15\)
- \(15 \div 3 = 5\) (5 is not divisible by 3 anymore)
- Finally, divide by 5:
- \(5 \div 5 = 1\)

2. **Counting the Factors**:
- From the factorization, we have:
- \(2\) appears \(4\) times (from \(2^4\))
- \(3\) appears \(3\) times (from \(3^3\))
- \(5\) appears \(1\) time (from \(5^1\))
- Therefore, we can express \(2160\) as:
\[ 2160 = 2^4 \cdot 3^3 \cdot 5^1 \]
- From this, we can identify:
- \(a = 4\)
- \(b = 3\)
- \(c = 1\)

3. **Calculating the Expression**:
- Now we need to calculate \(3^a \cdot 2^{-b} \cdot 5^{-c}\):
- Substitute the values of \(a\), \(b\), and \(c\):
\[ 3^a \cdot 2^{-b} \cdot 5^{-c} = 3^4 \cdot 2^{-3} \cdot 5^{-1} \]
- Calculate each part:
- \(3^4 = 81\)
- \(2^{-3} = \frac{1}{2^3} = \frac{1}{8}\)
- \(5^{-1} = \frac{1}{5}\)
- Combine these:
\[ 3^4 \cdot 2^{-3} \cdot 5^{-1} = 81 \cdot \frac{1}{8} \cdot \frac{1}{5} = \frac{81}{8 \cdot 5} = \frac{81}{40} \]

### Final Answer:
Thus, the value of \(3^a \cdot 2^{-b} \cdot 5^{-c}\) is:
\[ \frac{81}{40} \]