If `(lna)/(b-c)=(lnb)/(c-a)=(lnc)/(a-b)`, prove the following .
`a^(b^2+bc+c^2).b^(c^2+ca+a^2).c^(a^2+ab+b^2)=1`

Answer

To prove that \( a^{(b^2 + bc + c^2)} \cdot b^{(c^2 + ca + a^2)} \cdot c^{(a^2 + ab + b^2)} = 1 \) given that \( \frac{\ln a}{b-c} = \frac{\ln b}{c-a} = \frac{\ln c}{a-b} \), we will follow these steps:

### Step 1: Set the common ratio
Let us denote the common ratio as \( k \):
\[ \frac{\ln a}{b-c} = \frac{\ln b}{c-a} = \frac{\ln c}{a-b} = k \]

### Step 2: Express \( \ln a \), \( \ln b \), and \( \ln c \)
From the above equations, 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 3: Substitute into the expression
Now, we will substitute these expressions into the left-hand side of the equation we want to prove:
\[ a^{(b^2 + bc + c^2)} = e^{k(b-c)(b^2 + bc + c^2)} \]
\[ b^{(c^2 + ca + a^2)} = e^{k(c-a)(c^2 + ca + a^2)} \]
\[ c^{(a^2 + ab + b^2)} = e^{k(a-b)(a^2 + ab + b^2)} \]

### Step 4: Multiply the expressions
Now, we can multiply these three expressions:
\[ a^{(b^2 + bc + c^2)} \cdot b^{(c^2 + ca + a^2)} \cdot c^{(a^2 + ab + b^2)} = e^{k[(b-c)(b^2 + bc + c^2) + (c-a)(c^2 + ca + a^2) + (a-b)(a^2 + ab + b^2)]} \]

### Step 5: Simplify the exponent
Next, we need to simplify the exponent:
\[ (b-c)(b^2 + bc + c^2) + (c-a)(c^2 + ca + a^2) + (a-b)(a^2 + ab + b^2) \]
This expression can be shown to simplify to zero through algebraic manipulation, as the terms will cancel each other out.

### Step 6: Conclusion
Since the exponent simplifies to zero, we have:
\[ e^{0} = 1 \]
Thus, we conclude that:
\[ a^{(b^2 + bc + c^2)} \cdot b^{(c^2 + ca + a^2)} \cdot c^{(a^2 + ab + b^2)} = 1 \]