The value of $$ \left(x^{\frac{b+c}{c-a}}\right)^{\frac{1}{a-b}} \cdot \left(x^{\frac{c+a}{a-b}}\right)^{\frac{1}{b-c}} \cdot \left(x^{\frac{a+b}{b-c}}\right)^{\frac{1}{c-a}} $$ is

### Concept & Formula When raising a power to a power, multiply the exponents: $ (x^m)^n = x^{mn} $. When multiplying terms with the same base, add their exponents: $ x^m \cdot x^n = x^{m+n} $. For cyclic symmetry problems in competitive exams, the sum of the exponents almost always evaluates to zero. ### Step-by-Step Solution * **Step 1:** Apply the power-to-a-power rule to simplify each term's exponent. First term exponent = $ \frac{b+c}{(c-a)(a-b)} $ Second term exponent = $ \frac{c+a}{(a-b)(b-c)} $ Third term exponent = $ \frac{a+b}{(b-c)(c-a)} $ * **Step 2:** Multiply the bases by adding all three exponents together. Total Exponent = $ \frac{b+c}{(c-a)(a-b)} + \frac{c+a}{(a-b)(b-c)} + \frac{a+b}{(b-c)(c-a)} $ * **Step 3:** Take the Least Common Multiple (LCM) for the denominators, which is $ (a-b)(b-c)(c-a) $. Adjust the numerators accordingly: First term numerator = $ (b+c)(b-c) = b^2 - c^2 $ Second term numerator = $ (c+a)(c-a) = c^2 - a^2 $ Third term numerator = $ (a+b)(a-b) = a^2 - b^2 $ * **Step 4:** Sum the numerators. $ (b^2 - c^2) + (c^2 - a^2) + (a^2 - b^2) = 0 $ * **Step 5:** Evaluate the final base and exponent. Since the numerator of our exponent is $0$, the total exponent is $0$. $ x^0 = 1 $ (assuming $x \neq 0$) ### Exam Strategy & Shortcut **Pattern Recognition:** Cyclic equations (where variables shift in a continuous loop like $a \rightarrow b \rightarrow c$) usually simplify symmetrically. In exponent problems of this exact structure, the exponents invariably cancel out to yield $x^0 = 1$. **Value Substitution:** If you forget the algebra, assume $a=1, b=2, c=3$. (Never pick values that make a denominator zero). The arithmetic will quickly cancel down to $x^0 = 1$, allowing you to mark the answer in under 15 seconds. ### Common Pitfall A major trap is making sign errors when establishing the common denominator. For example, treating $(c-a)$ as $-(a-c)$ and losing track of the negative sign will prevent the numerators from perfectly canceling out. ### Final Answer Therefore, the correct answer is **1**.