Differentiate `((x^(a))/(x^(b)))^(a+b)*((x^(b))/(x^(c)))^(b+c)*((x^(c))/(x^(a)))^(c+a)` with respect to x.

Answer

To differentiate the function

\[ f(x) = \left(\frac{x^a}{x^b}\right)^{a+b} \cdot \left(\frac{x^b}{x^c}\right)^{b+c} \cdot \left(\frac{x^c}{x^a}\right)^{c+a} \]

with respect to \(x\), we can simplify it first before differentiating.

### Step-by-step Solution:

1. **Simplify each term:**
We can rewrite each fraction:
\[ \frac{x^a}{x^b} = x^{a-b}, \quad \frac{x^b}{x^c} = x^{b-c}, \quad \frac{x^c}{x^a} = x^{c-a} \]
Therefore, we can express \(f(x)\) as:
\[ f(x) = \left(x^{a-b}\right)^{a+b} \cdot \left(x^{b-c}\right)^{b+c} \cdot \left(x^{c-a}\right)^{c+a} \]

2. **Apply the power rule:**
Using the power of a power property \((x^m)^n = x^{mn}\), we can rewrite \(f(x)\):
\[ f(x) = x^{(a-b)(a+b)} \cdot x^{(b-c)(b+c)} \cdot x^{(c-a)(c+a)} \]

3. **Combine the exponents:**
Since the bases are the same, we can add the exponents:
\[ f(x) = x^{(a-b)(a+b) + (b-c)(b+c) + (c-a)(c+a)} \]

4. **Simplify the exponent:**
Let's denote the exponent as:
\[ E = (a-b)(a+b) + (b-c)(b+c) + (c-a)(c+a) \]
Expanding each term:
\[ E = (a^2 - b^2) + (b^2 - c^2) + (c^2 - a^2) \]
Notice that \(a^2 - b^2 + b^2 - c^2 + c^2 - a^2 = 0\). Hence:
\[ E = 0 \]

5. **Final expression for \(f(x)\):**
Therefore, we have:
\[ f(x) = x^0 = 1 \]

6. **Differentiate \(f(x)\):**
Now, we differentiate \(f(x)\) with respect to \(x\):
\[ f'(x) = \frac{d}{dx}(1) = 0 \]

### Final Answer:
Thus, the derivative of the function is:
\[ f'(x) = 0 \]