If `(Ina)/(b-c)=(Inb)/(c-a)=(Inc)/(a-b)`, prove the following .
abc=1

Answer

To prove that \( abc = 1 \) given that

\[ \frac{\ln a}{b - c} = \frac{\ln b}{c - a} = \frac{\ln c}{a - b} = k, \]

we can follow these steps:

### Step 1: Set up the equations
From the given equality, we can express \( \ln a \), \( \ln b \), and \( \ln c \) in terms of \( k \):

\[ \ln a = k(b - c), \]
\[ \ln b = k(c - a), \]
\[ \ln c = k(a - b). \]

### Step 2: Exponentiate the equations
Next, we exponentiate each equation to express \( a \), \( b \), and \( c \):

\[ a = e^{k(b - c)}, \]
\[ b = e^{k(c - a)}, \]
\[ c = e^{k(a - b)}. \]

### Step 3: Multiply the equations
Now, we multiply \( a \), \( b \), and \( c \):

\[ abc = e^{k(b - c)} \cdot e^{k(c - a)} \cdot e^{k(a - b)}. \]

### Step 4: Combine the exponents
Since the bases are the same, we can combine the exponents:

\[ abc = e^{k[(b - c) + (c - a) + (a - b)]}. \]

### Step 5: Simplify the exponent
Now, simplify the exponent:

\[ (b - c) + (c - a) + (a - b) = 0. \]

Thus, we have:

\[ abc = e^{k \cdot 0} = e^0 = 1. \]

### Conclusion
Therefore, we conclude that:

\[ abc = 1. \]