If `loga/(b-c) = logb/(c-a) = logc/(a-b)`, then `a^(b+c).b^(c+a).c^(a+b)`=

Options

A. `a^(b)b^(c)c^(a)=1`
B. `a^(a)b^(b)c^(c)=1`
C. `root(a)(a) root(b)(b) root(c)(c) = 1`
D. none of these

Explanation

To solve the problem, we start with the given equation:

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

### Step 1: Express logs in terms of \( k \)

From the above equation, we can express the logarithms of \( a \), \( b \), and \( c \) in terms of \( k \):

\[ \log a = k(b - c) \quad (1) \]
\[ \log b = k(c - a) \quad (2) \]
\[ \log c = k(a - b) \quad (3) \]

### Step 2: Add the equations

Now, we add equations (1), (2), and (3):

\[ \log a + \log b + \log c = k(b - c + c - a + a - b) \]

The terms inside the parentheses cancel out:

\[ \log a + \log b + \log c = k(0) = 0 \]

### Step 3: Use the property of logarithms

Since \( \log a + \log b + \log c = 0 \), we can use the property of logarithms:

\[ \log(abc) = 0 \]

### Step 4: Take the antilogarithm

Taking the antilogarithm of both sides gives us:

\[ abc = e^0 = 1 \]

### Step 5: Find the expression \( a^{(b+c)} \cdot b^{(c+a)} \cdot c^{(a+b)} \)

We need to find the value of:

\[ a^{(b+c)} \cdot b^{(c+a)} \cdot c^{(a+b)} \]

### Step 6: Rewrite the expression

We can rewrite the expression as follows:

\[ = a^b \cdot a^c \cdot b^c \cdot b^a \cdot c^a \cdot c^b \]

This can be rearranged as:

\[ = a^a \cdot b^b \cdot c^c \cdot a^{b+c} \cdot b^{c+a} \cdot c^{a+b} \]

### Step 7: Introduce \( abc \)

Now we can multiply and divide by \( a^a \cdot b^b \cdot c^c \):

\[ = \frac{a^{(b+c)} \cdot b^{(c+a)} \cdot c^{(a+b)} \cdot a^a \cdot b^b \cdot c^c}{a^a \cdot b^b \cdot c^c} \]

### Step 8: Simplify

Since \( abc = 1 \), we have:

\[ = \frac{(abc)^{(a+b+c)}}{abc} = \frac{1^{(a+b+c)}}{1} = 1 \]

### Conclusion

Thus, the final result is:

\[ a^{(b+c)} \cdot b^{(c+a)} \cdot c^{(a+b)} = 1 \]