(x^a/x^-b)^a-b × (x^b/x^-c)^b-c × (x^c/x^-a)^c-a = 1

1.	(x^a/x^-b)^a-b × (x^b/x^-c)^b-c × (x^c/x^-a)^c-a = 1
Answer» PROVE That $\sf{ \left( \cfrac{ {x}^{a} }{ {x}^{ - b} }\right)^{a - b} \times\left( \cfrac{ {x}^{b} }{ {x}^{ - c} }\right)^{b- c} \times\left( \cfrac{ {x}^{c} }{ {x}^{ - a} }\right)^{c - a} = 1}$ Solution Let $\sf{<klux>LHS</klux>= \left( \cfrac{ {x}^{a} }{ {x}^{ - b} }\right)^{a - b} \times\left( \cfrac{ {x}^{b} }{ {x}^{ - c} }\right)^{b- c} \times\left( \cfrac{ {x}^{c} }{ {x}^{ - a} }\right)^{c - a} }$ $\text{[math]}$ On SOLVING LHS we get $\sf{LHS= \left( \cfrac{ {x}^{a} }{ {x}^{ - b} }\right)^{a - b} \times\left( \cfrac{ {x}^{b} }{ {x}^{ - c} }\right)^{b- c} \times\left( \cfrac{ {x}^{c} }{ {x}^{ - a} }\right)^{c - a} } \\ \\ \sf{ = \left( \cfrac{ {x}^{ {a}^{2} - ab} }{ {x}^{ { - b}^{2} - ab } }\right) \times \left( \cfrac{ {x}^{ {b}^{2} - bc} }{ {x}^{ { - c}^{2} - bc} }\right) \times \left( \cfrac{ {x}^{ {c}^{2} - ca} }{ {x}^{ { - a}^{2} - ca } }\right) } \\ \\ \sf{ = \left( \cfrac{ {x}^{ {a}^{2} } \times {x}^{ - ab} }{ {x}^{ { b}^{2} } \times {x}^{ - ab} }\right) \times \left( \cfrac{ {x}^{ {b}^{2} } \times {x}^{ - bc} }{ {x}^{ { c}^{2} } \times {x}^{ - bc} }\right)\times \left( \cfrac{ {x}^{ {c}^{2} } \times {x}^{ - ca} }{ {x}^{ { a}^{2} } \times {x}^{ - ca} }\right)} \\ \\ { = \sf \frac{ {x}^{ {a}^{2} } }{ {x}^{ {b}^{2} } } \times \sf \frac{ {x}^{ {b}^{2} } }{ {x}^{ {c}^{2} } } \times \sf \frac{ {x}^{ {c}^{2} } }{ {x}^{ {a}^{2} } } } \\ \\ { = \sf \frac{ \xcancel{ \: \: \: \:{x}^{ {a}^{2} }} }{ \xcancel{ \: \: \: \: {x}^{ {b}^{2} } } } \times \sf \frac{ \xcancel{ \: \: \: \: {x}^{ {b}^{2} }} }{ \xcancel{ \: \: \: \: {x}^{ {c}^{2} } }} \times \sf \frac{ \xcancel{ \: \: \: \: {x}^{ {c}^{2} }} }{ \xcancel{{x}^{ {a}^{2} } } } } \\ \\ = 1$ $\sf RHS=1$ LHS=RHS PROVED

Answer»

PROVE That

$\sf{ \left( \cfrac{ {x}^{a} }{ {x}^{ - b} }\right)^{a - b} \times\left( \cfrac{ {x}^{b} }{ {x}^{ - c} }\right)^{b- c} \times\left( \cfrac{ {x}^{c} }{ {x}^{ - a} }\right)^{c - a} = 1}$

Solution

Let

$\sf{<klux>LHS</klux>= \left( \cfrac{ {x}^{a} }{ {x}^{ - b} }\right)^{a - b} \times\left( \cfrac{ {x}^{b} }{ {x}^{ - c} }\right)^{b- c} \times\left( \cfrac{ {x}^{c} }{ {x}^{ - a} }\right)^{c - a} }$

$\text{[math]}$

On SOLVING LHS we get

$\sf{LHS= \left( \cfrac{ {x}^{a} }{ {x}^{ - b} }\right)^{a - b} \times\left( \cfrac{ {x}^{b} }{ {x}^{ - c} }\right)^{b- c} \times\left( \cfrac{ {x}^{c} }{ {x}^{ - a} }\right)^{c - a} } \\ \\ \sf{ = \left( \cfrac{ {x}^{ {a}^{2} - ab} }{ {x}^{ { - b}^{2} - ab } }\right) \times \left( \cfrac{ {x}^{ {b}^{2} - bc} }{ {x}^{ { - c}^{2} - bc} }\right) \times \left( \cfrac{ {x}^{ {c}^{2} - ca} }{ {x}^{ { - a}^{2} - ca } }\right) } \\ \\ \sf{ = \left( \cfrac{ {x}^{ {a}^{2} } \times {x}^{ - ab} }{ {x}^{ { b}^{2} } \times {x}^{ - ab} }\right) \times \left( \cfrac{ {x}^{ {b}^{2} } \times {x}^{ - bc} }{ {x}^{ { c}^{2} } \times {x}^{ - bc} }\right)\times \left( \cfrac{ {x}^{ {c}^{2} } \times {x}^{ - ca} }{ {x}^{ { a}^{2} } \times {x}^{ - ca} }\right)} \\ \\ { = \sf \frac{ {x}^{ {a}^{2} } }{ {x}^{ {b}^{2} } } \times \sf \frac{ {x}^{ {b}^{2} } }{ {x}^{ {c}^{2} } } \times \sf \frac{ {x}^{ {c}^{2} } }{ {x}^{ {a}^{2} } } } \\ \\ { = \sf \frac{ \xcancel{ \: \: \: \:{x}^{ {a}^{2} }} }{ \xcancel{ \: \: \: \: {x}^{ {b}^{2} } } } \times \sf \frac{ \xcancel{ \: \: \: \: {x}^{ {b}^{2} }} }{ \xcancel{ \: \: \: \: {x}^{ {c}^{2} } }} \times \sf \frac{ \xcancel{ \: \: \: \: {x}^{ {c}^{2} }} }{ \xcancel{{x}^{ {a}^{2} } } } } \\ \\ = 1$

$\sf RHS=1$

LHS=RHS PROVED