Answer:
[tex]\sf \dfrac{{x}^{a \: (b - c)} }{ {x}^{b \: (a - c)} } \: \div \: \bigg \lgroup \dfrac{ {x}^{b} }{ {x}^{a} } \bigg \rgroup^{c} \: = \: 1[/tex]
[tex]\\[/tex]
Now,
[tex]\sf \dfrac{{x}^{ab \: - \: ac} }{ {x}^{ba \: - \: b c} } \: \div \: \bigg \lgroup \dfrac{ ({x}^{b} )}{( {x}^{a}) } \bigg \rgroup^{c} \: = \: 1[/tex]
[tex]\\[/tex]
[tex]\sf \dfrac{{x}^{ab \: - \: ac} }{ {x}^{ba \: - \: b c} } \: \div \: \dfrac{ {x}^{bc} }{{x}^{ac} } \: = \: 1[/tex]
[tex]\\[/tex]
We know that, When sign change from division to multiplication we should do reciprocal of next number.
Therefore we get,
[tex]\sf \dfrac{{x}^{ab \: - \: ac} }{ {x}^{ba \: - \: b c} } \: \times \: \dfrac{ {x}^{ac} }{{x}^{bc} } \: = \: 1[/tex]
[tex]\\[/tex]
[tex]\sf \dfrac{{x}^{ab \: - \: \cancel {ac }\: + \: \cancel{ac}} }{ {x}^{ba \: - \: \cancel{ b c} \: + \: \cancel{bc}} } \: = \: 1[/tex]
[tex]\\[/tex]
[tex]\sf \dfrac{{x} \: ^{ \cancel{ab}} }{ {x} \: ^{\cancel{ba}} } \: = \: 1[/tex]
[tex]\\[/tex]
[tex] \bigstar \: \: \underline{ \boxed{\sf x \: = \: 1}} \: \: \bigstar[/tex]
[tex]\\[/tex]
[tex] \huge\bf \dag \: \gray{RHS = LHS} \: \dag[/tex]
[tex]\\[/tex]
[tex]\large \star \: \tt Hence, Verified \: \star[/tex]
