$${\left( {\frac{{{x^b}}}{{{x^c}}}} \right)^{\left( {b + c - a} \right)}}.$$   $${\left( {\frac{{{x^c}}}{{{x^a}}}} \right)^{\left( {c + a - b} \right)}}.$$   $${\left( {\frac{{{x^a}}}{{{x^b}}}} \right)^{\left( {a + b - c} \right)}}$$   = ?

Solution(By Examveda Team)

$$\eqalign{ & {\text{Given Exp}}{\text{.}} \cr & = {x^{\left( {b - c} \right)\left( {b + c - a} \right)}}.{x^{\left( {c - a} \right)\left( {c + a - b} \right)}}.{x^{\left( {a - b} \right)\left( {a + b - c} \right)}} \cr} $$
  $$ = {x^{\left( {b - c} \right)\left( {b + c} \right) - a\left( {b - c} \right)}}.$$    $${x^{\left( {c - a} \right)\left( {c + a} \right) - b\left( {c - a} \right)}}.$$   $${x^{\left( {a - b} \right)\left( {a + b} \right) - c\left( {a - b} \right)}}$$
$$\eqalign{ & = {x^{\left( {{b^2} - {c^2} + {c^2} - {a^2} + {a^2} - {b^2}} \right)}}.{x^{ - a\left( {b - c} \right) - b\left( {c - a} \right) - c\left( {a - b} \right)}} \cr & = {{x^0} . {x^\left( -ab + ac - bc + ba - ca + cb \right)}} \cr & = \left( {{x^0} \times {x^0}} \right) \cr & = \left( {1 \times 1} \right) \cr & = 1 \cr} $$