`((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))=?`

Options

A. `3`
B. `2`
C. `1`
D. `0`

Answer

To solve the expression

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

we can follow these steps:

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

\[ \frac{x^a}{x^b} = x^{a-b}, \quad \frac{x^b}{x^c} = x^{b-c}, \quad \frac{x^c}{x^a} = x^{c-a} \]

### Step 2: Rewrite the expression
Now we can rewrite the entire expression as:

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

### Step 3: Apply the power of a power property
Using the property \((x^m)^n = x^{mn}\), we can simplify each term:

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

### 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: Recognize the identity
Notice that the terms in the exponent resemble the identity for the difference of cubes:

\[ A^3 - B^3 = (A-B)(A^2 + AB + B^2) \]

In our case, we can identify:

- \(A = a\), \(B = b\)
- \(A = b\), \(B = c\)
- \(A = c\), \(B = a\)

This means that the expression simplifies to:

\[ x^{(a^3 - b^3) + (b^3 - c^3) + (c^3 - a^3)} \]

### Step 6: Simplify the exponent
When we add these terms, we see that:

\[ a^3 - b^3 + b^3 - c^3 + c^3 - a^3 = 0 \]

Thus, the exponent simplifies to \(0\).

### Step 7: Final result
Since \(x^0 = 1\) for any non-zero \(x\), we conclude that:

\[ \left(\frac{x^a}{x^b}\right)^{a^2 + ab + b^2} \left(\frac{x^b}{x^c}\right)^{b^2 + bc + c^2} \left(\frac{x^c}{x^a}\right)^{c^2 + ca + a^2} = 1 \]