The value `((x^a)/(x^b))^((a^2 + ab + b^2)) ((x^b)/(x^c))^((b^2 + bc + c^2))((x^c)/(x^a))^((c^2 + ca + a^2))` is :

Options

A. `-1`
B. `x^(abc)`
C. `1`
D. `x^(a +b + c)`

Answer

To solve the expression

\[ \left(\frac{x^a}{x^b}\right)^{(a^2 + ab + b^2)} \cdot \left(\frac{x^b}{x^c}\right)^{(b^2 + bc + c^2)} \cdot \left(\frac{x^c}{x^a}\right)^{(c^2 + ca + a^2)}, \]

we can follow these steps:

### Step 1: Simplify each fraction
Using the property of exponents, \(\frac{x^m}{x^n} = x^{m-n}\), we can simplify each fraction in the expression:

1. \(\frac{x^a}{x^b} = x^{a-b}\)
2. \(\frac{x^b}{x^c} = x^{b-c}\)
3. \(\frac{x^c}{x^a} = x^{c-a}\)

### Step 2: Substitute back into the expression
Substituting these simplifications back into the expression gives us:

\[ \left(x^{a-b}\right)^{(a^2 + ab + b^2)} \cdot \left(x^{b-c}\right)^{(b^2 + bc + c^2)} \cdot \left(x^{c-a}\right)^{(c^2 + ca + a^2)}. \]

### Step 3: Apply the power of a power property
Using the property \((x^m)^n = x^{m \cdot n}\), we can rewrite the expression as:

\[ x^{(a-b)(a^2 + ab + b^2)} \cdot x^{(b-c)(b^2 + bc + c^2)} \cdot x^{(c-a)(c^2 + ca + a^2)}. \]

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

\[ x^{(a-b)(a^2 + ab + b^2) + (b-c)(b^2 + bc + c^2) + (c-a)(c^2 + ca + a^2)}. \]

### Step 5: Analyze the exponent
Now we need to analyze the exponent:

\[ E = (a-b)(a^2 + ab + b^2) + (b-c)(b^2 + bc + c^2) + (c-a)(c^2 + ca + a^2). \]

This expression can be recognized as a symmetric polynomial, and it can be shown that it simplifies to zero. This is because each term represents a cyclic permutation of \(a\), \(b\), and \(c\) and will cancel out.

### Step 6: Conclusion
Since the exponent \(E\) simplifies to zero, we have:

\[ x^E = x^0 = 1. \]

Thus, the value of the original expression is:

\[ \boxed{1}. \]