Question

show that
(x^a/x^-b) ^a-b. (x^b/x^-c)^b-c. (x^c/x^-a) ^c-a=1

​

Answer 1

Question:

Show that:

${\large{\rm {{( \dfrac{ {x}^{a} }{ {x}^{ - b} } )}^{a - b} .{( \dfrac{ {x}^{b} }{ {x}^{ - c} } )}^{b - c}.{( \dfrac{ {x}^{c} }{ {x}^{ - a} } )}^{c - a} = 1}}}$

Solution:

We are given,

${\large{\rm {{( \dfrac{ {x}^{a} }{ {x}^{ - b} } )}^{a - b} .{( \dfrac{ {x}^{b} }{ {x}^{ - c} } )}^{b - c}.{( \dfrac{ {x}^{c} }{ {x}^{ - a} } )}^{c - a} = 1}}}$

here ,

LHS =

${\rm{ {( \dfrac{ {x}^{a} }{ {x}^{ - b} } )}^{a - b} .{( \dfrac{ {x}^{b} }{ {x}^{ - c} } )}^{b - c}.{( \dfrac{ {x}^{c} }{ {x}^{ - a} } )}^{c - a}}}$

we know,

${\implies{\red{\boxed{\green{\star{\rm{ \dfrac{ {m}^{a} }{ {m}^{ b} } = {m}^{a - b}} }}}}}}$

${\longrightarrow{\rm{ { ({x}^{a - ( - b)}) }^{a - b} .{ ({x}^{b- ( - c)}) }^{b - c} {( {x}^{c - ( - a)}) }^{c - a} }}}$

${\longrightarrow{\rm{{ ({x}^{a + b}) }^{a - b} .{ ({x}^{b + c}) }^{b - c} {( {x}^{c + a}) }^{c - a} }}}$

we also know,

${\implies{\red{\boxed{\green{\star{\rm{{ ({m}^{a}) }^{b} = {m}^{ab}}}}}}}}$

${\longrightarrow{\rm{ { {x}^{(a + b)(a - b)} }. {x}^{(b + c)(b - c)} . {x}^{(c + a)(c - a)}}}}$

by using,

${\implies{\red{\boxed{\green{\star{\rm{(m + n)(m - n) = {m}^{2} - {n}^{2} }}}}}}}$

${\longrightarrow{\rm{ {x}^{ {a}^{2} - {b}^{2} } .{x}^{ {b}^{2} - {c}^{2} } .{x}^{ {c}^{2} - {a}^{2} }}}}$

but,

${\implies{\red{\boxed{\green{\star{\rm{{m}^{a} . {m}^{b} = {m}^{a + b}}}}}}}}$

${\implies{\rm{{x}^{( {a}^{2} - {b}^{2} ) + ( {b}^{2} - {c}^{2} ) + ( {c}^{2} - {a}^{2} )}}}}$

${\implies{\rm{ {x}^{( {a}^{2 } - {b}^{2} + {b}^{2} - {c}^{2} + {c}^{2} - {a}^{2}) }}}}$

${\implies{\rm{{x}^{0}}}}$

${\implies{\rm{1}}}$

= RHS

${\therefore{\orange{\boxed{\blue{\star{ {( \frac{ {x}^{a} }{ {x}^{ - b} } )}^{a - b} .{( \frac{ {x}^{b} }{ {x}^{ - c} } )}^{b - c}.{( \frac{ {x}^{c} }{ {x}^{ - a} } )}^{c - a} = 1 }}}}}}$

Hence proved .