Question
$${\left( {\frac{{{x^b}}}{{{x^c}}}} \right)^{\left( {b + c - a} \right)}}.$$ $${\left( {\frac{{{x^c}}}{{{x^a}}}} \right)^{\left( {c + a - b} \right)}}.$$ $${\left( {\frac{{{x^a}}}{{{x^b}}}} \right)^{\left( {a + b - c} \right)}} = ?$$
Answer: Option B
$${x^{\left( {b - c} \right)\left( {b + c - a} \right)}}.{x^{\left( {c - a} \right)\left( {c + a - b} \right)}}.{x^{\left( {a - b} \right)\left( {a + b - c} \right)}}$$
$$ = {x^{\left( {b - c} \right)\left( {b + c} \right) - a\left( {b - c} \right)}}.$$ $${x^{\left( {c - a} \right)\left( {c + a} \right) - b\left( {c - a} \right)}}.$$ $${x^{\left( {a - b} \right)\left( {a + b} \right) - c\left( {a - b} \right)}}$$
$$\eqalign{
& = {x^{\left( {{b^2} - {c^2} + {c^2} - {a^2} + {a^2} - {b^2}} \right)}}.{x^{ - a\left( {b - c} \right) - b\left( {c - a} \right) - c\left( {a - b} \right)}} \cr
& = \left( {{x^0} \times {x^0}} \right) \cr
& = \left( {1 \times 1} \right) \cr
& = 1 \cr} $$
Was this answer helpful ?
$${x^{\left( {b - c} \right)\left( {b + c - a} \right)}}.{x^{\left( {c - a} \right)\left( {c + a - b} \right)}}.{x^{\left( {a - b} \right)\left( {a + b - c} \right)}}$$
$$ = {x^{\left( {b - c} \right)\left( {b + c} \right) - a\left( {b - c} \right)}}.$$ $${x^{\left( {c - a} \right)\left( {c + a} \right) - b\left( {c - a} \right)}}.$$ $${x^{\left( {a - b} \right)\left( {a + b} \right) - c\left( {a - b} \right)}}$$
$$\eqalign{
& = {x^{\left( {{b^2} - {c^2} + {c^2} - {a^2} + {a^2} - {b^2}} \right)}}.{x^{ - a\left( {b - c} \right) - b\left( {c - a} \right) - c\left( {a - b} \right)}} \cr
& = \left( {{x^0} \times {x^0}} \right) \cr
& = \left( {1 \times 1} \right) \cr
& = 1 \cr} $$
Was this answer helpful ?
More Questions on This Topic :
Latest Videos
Latest Test Papers
Submit Solution