If `(logx)/(a^2+a b+b^2)=(logy)/(b^2+b c+c^2)=(logz)/(c^2+c a+a^2),` then `x^(a-b)*y^(b-c)*z^(c-a)=`

Options

A. 0
B. `-1`
C. 1
D. 2

Answer

To solve the equation

\[ \frac{\log x}{a^2 + ab + b^2} = \frac{\log y}{b^2 + bc + c^2} = \frac{\log z}{c^2 + ca + a^2}, \]

we will denote the common value of these ratios as \( k \). Thus, we can write:

1. \(\log x = k(a^2 + ab + b^2)\)
2. \(\log y = k(b^2 + bc + c^2)\)
3. \(\log z = k(c^2 + ca + a^2)\)

Next, we will manipulate each equation to express \( x^{a-b} \), \( y^{b-c} \), and \( z^{c-a} \).

### Step 1: Express \( x^{a-b} \)

From the first equation:

\[ \log x = k(a^2 + ab + b^2) \]

Multiplying both sides by \( a - b \):

\[ (a - b) \log x = k(a - b)(a^2 + ab + b^2) \]

Using the property of logarithms \( n \log m = \log m^n \):

\[ \log(x^{a-b}) = k(a - b)(a^2 + ab + b^2) \]

Thus, we can express \( x^{a-b} \) as:

\[ x^{a-b} = 10^{k(a - b)(a^2 + ab + b^2)} \]

### Step 2: Express \( y^{b-c} \)

From the second equation:

\[ \log y = k(b^2 + bc + c^2) \]

Multiplying both sides by \( b - c \):

\[ (b - c) \log y = k(b - c)(b^2 + bc + c^2) \]

Using the logarithmic property again:

\[ \log(y^{b-c}) = k(b - c)(b^2 + bc + c^2) \]

Thus, we can express \( y^{b-c} \) as:

\[ y^{b-c} = 10^{k(b - c)(b^2 + bc + c^2)} \]

### Step 3: Express \( z^{c-a} \)

From the third equation:

\[ \log z = k(c^2 + ca + a^2) \]

Multiplying both sides by \( c - a \):

\[ (c - a) \log z = k(c - a)(c^2 + ca + a^2) \]

Using the logarithmic property again:

\[ \log(z^{c-a}) = k(c - a)(c^2 + ca + a^2) \]

Thus, we can express \( z^{c-a} \) as:

\[ z^{c-a} = 10^{k(c - a)(c^2 + ca + a^2)} \]

### Step 4: Combine the results

Now, we need to find the product:

\[ x^{a-b} \cdot y^{b-c} \cdot z^{c-a} \]

Substituting the expressions we derived:

\[ x^{a-b} \cdot y^{b-c} \cdot z^{c-a} = 10^{k(a - b)(a^2 + ab + b^2)} \cdot 10^{k(b - c)(b^2 + bc + c^2)} \cdot 10^{k(c - a)(c^2 + ca + a^2)} \]

Combining the exponents:

\[ = 10^{k\left((a - b)(a^2 + ab + b^2) + (b - c)(b^2 + bc + c^2) + (c - a)(c^2 + ca + a^2)\right)} \]

### Step 5: Simplifying the exponent

Notice that:

1. The terms \( a^2 + ab + b^2 \), \( b^2 + bc + c^2 \), and \( c^2 + ca + a^2 \) will cancel out when we expand the expression because each term appears once positively and once negatively in the overall sum.

Thus, the entire exponent simplifies to zero:

\[ = 10^0 = 1 \]

### Final Answer

Therefore, the value of \( x^{a-b} \cdot y^{b-c} \cdot z^{c-a} \) is:

\[ \boxed{1} \]