The value of `((x^(a^2) )/( x^(b^2) ) )^( ( 1)/( a+b) ) . ((x^(b^2) )/(x^( c^2) ) )^( (1)/( b+ c) ) . ((x^( c^2) )/( x^( a^2) ) )^( (1)/( c+a) ) ` is

Options

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

Answer

To solve the expression

\[ \left( \frac{x^{a^2}}{x^{b^2}} \right)^{\frac{1}{a+b}} \cdot \left( \frac{x^{b^2}}{x^{c^2}} \right)^{\frac{1}{b+c}} \cdot \left( \frac{x^{c^2}}{x^{a^2}} \right)^{\frac{1}{c+a}} \]

we will simplify it step by step.

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

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

### Step 2: Rewrite the Expression
Now we can rewrite the entire expression using the simplified fractions:

\[ \left( x^{a^2 - b^2} \right)^{\frac{1}{a+b}} \cdot \left( x^{b^2 - c^2} \right)^{\frac{1}{b+c}} \cdot \left( x^{c^2 - a^2} \right)^{\frac{1}{c+a}} \]

### Step 3: Apply the Power of a Power Rule
Using the property of exponents that states \((x^m)^n = x^{m \cdot n}\), we can simplify further:

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

### Step 4: Combine the Exponents
Now we can combine the exponents since the bases are the same:

\[ x^{\frac{a^2 - b^2}{a+b} + \frac{b^2 - c^2}{b+c} + \frac{c^2 - a^2}{c+a}} \]

### Step 5: Simplify the Combined Exponent
Now we will simplify the exponent:

- The term \(\frac{a^2 - b^2}{a+b}\) can be factored using the identity \(a^2 - b^2 = (a-b)(a+b)\) to give \(a - b\).
- Similarly, \(\frac{b^2 - c^2}{b+c} = b - c\) and \(\frac{c^2 - a^2}{c+a} = c - a\).

Thus, we have:

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

### Step 6: Combine Like Terms
Now, combine the terms in the exponent:

\[ (a - b) + (b - c) + (c - a) = a - b + b - c + c - a = 0 \]

### Step 7: Final Result
Since the exponent simplifies to 0, we have:

\[ x^0 = 1 \]

Thus, the value of the original expression is:

\[ \boxed{1} \]