Question
The value of $${\left( {{x^{\frac{{b + c}}{{c - a}}}}} \right)^{\frac{1}{{a - b}}}}{\text{.}}$$ $${\left( {{x^{\frac{{c + a}}{{a - b}}}}} \right)^{\frac{1}{{b - c}}}}.$$ $${\left( {{x^{\frac{{a + b}}{{b - c}}}}} \right)^{\frac{1}{{c - a}}}}{\text{ is = ?}}$$
Answer: Option A
$$\eqalign{
& {x^{\frac{{b + c}}{{\left( {a - b} \right)\left( {c - a} \right)}}}}.{x^{\frac{{c + a}}{{\left( {a - b} \right)\left( {b - c} \right)}}}}.{x^{\frac{{a + b}}{{\left( {b - c} \right)\left( {c - a} \right)}}}} \cr
& = {x^{\frac{{\left( {b + c} \right)\left( {b - c} \right) + \left( {c + a} \right)\left( {c - a} \right) + \left( {a + b} \right)\left( {a - b} \right)}}{{\left( {a - b} \right)\left( {b - c} \right)\left( {c - a} \right)}}}} \cr
& = {x^{\frac{{\left( {{b^2} - {c^2}} \right) + \left( {{c^2} - {a^2}} \right) + \left( {{a^2} - {b^2}} \right)}}{{\left( {a - b} \right)\left( {b - c} \right)\left( {c - a} \right)}}}} \cr
& = {x^0} \cr
& = 1 \cr} $$
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$$\eqalign{
& {x^{\frac{{b + c}}{{\left( {a - b} \right)\left( {c - a} \right)}}}}.{x^{\frac{{c + a}}{{\left( {a - b} \right)\left( {b - c} \right)}}}}.{x^{\frac{{a + b}}{{\left( {b - c} \right)\left( {c - a} \right)}}}} \cr
& = {x^{\frac{{\left( {b + c} \right)\left( {b - c} \right) + \left( {c + a} \right)\left( {c - a} \right) + \left( {a + b} \right)\left( {a - b} \right)}}{{\left( {a - b} \right)\left( {b - c} \right)\left( {c - a} \right)}}}} \cr
& = {x^{\frac{{\left( {{b^2} - {c^2}} \right) + \left( {{c^2} - {a^2}} \right) + \left( {{a^2} - {b^2}} \right)}}{{\left( {a - b} \right)\left( {b - c} \right)\left( {c - a} \right)}}}} \cr
& = {x^0} \cr
& = 1 \cr} $$
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