The value of `(x^(a-b))^(c ) xx (x^(b-c))^(a) xx (x^(c-a))^(b)` is

Options

A. 0
B. 1
C. `x^(ab)`
D. `x^(bc)`

Answer

To solve the expression \((x^{(a-b)})^{c} \cdot (x^{(b-c)})^{a} \cdot (x^{(c-a)})^{b}\), we will use the properties of exponents. Here’s the step-by-step solution:

### Step 1: Apply the Power of a Power Property
Using the property \((x^m)^n = x^{m \cdot n}\), we can rewrite each term:

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

### Step 2: Combine the Exponents
Now, we can combine these exponents since they have the same base \(x\):

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

### Step 3: Simplify the Exponent
Next, we simplify the exponent:

\[ (a-b)c + (b-c)a + (c-a)b \]

Expanding each term:

- The first term: \(ac - bc\)
- The second term: \(ab - ac\)
- The third term: \(bc - ab\)

Now, combine these:

\[ (ac - bc) + (ab - ac) + (bc - ab) \]

### Step 4: Combine Like Terms
Now, combine the like terms:

- \(ac - ac\) cancels out
- \(ab - ab\) cancels out
- \(-bc + bc\) cancels out

Thus, we have:

\[ 0 \]

### Step 5: Final Result
So, we can conclude:

\[ x^{0} = 1 \]

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

\[ \boxed{1} \]