Answer:
[tex] \huge{ \boxed{ \tt{1 }}}[/tex]
❁ Question : Simplify :
- [tex] \sf{( {x}^{a} ) ^{b - c} \times ( {x}^{b} ) ^{c - a} \times ( {x}^{c} )^{a - b}} [/tex]
❁ Solution :
First , Use power law of indices.
Remember : If [tex] \sf{ {a}^{m} }[/tex] is an algebraic term , then
[tex] \sf{( {a}^{m} ) ^{n} } = {a}^{m \times n} = {a}^{mn} [/tex] , where m and n are positive integers.
➝ [tex] \sf{ {x}^{a(b - c)} \times {x}^{b(c - a)} \times {x}^{c(a - b)}} [/tex]
➝ [tex] \sf{ {x}^{ab - ac} \times {x}^{bc - ba} \times {x}^{ca - cb}} [/tex]
Now , Use product law of indices :
Remember : If [tex] \sf{ {a}^{m} }[/tex] and [tex] \sf{ {a}^{n}} [/tex] are the two algebraic terms , where m and n are the positive integers then [tex] \sf{ {a}^{m} \times {a}^{n} = {a}^{m + n}} [/tex]
➝ [tex] \sf{ {x}^{ab - ac + bc - ba + ca - cb}} [/tex]
Since two opposites adds up to zero , remove them :
➝ [tex] \sf{ {x} \: ^{ \cancel{ab} \: - \cancel{ac} \: + \cancel{bc} \: - \cancel{ba} \: + \cancel{ca} \: - \cancel{cb} } }[/tex]
➝ [tex] \sf{ {x}^{0} }[/tex]
Use Law of zero index
Remember : If [tex] \sf{ {a}^{0}} [/tex] is an algebraic term , where a ≠ 0 , then [tex] \sf{ {a}^{0} = 1}[/tex]
➝ [tex] \boxed{ \sf{1}}[/tex]
And we're done !
Hope I helped ! ♡
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