Question

ITS VERY URGENT........... .. . ..................(x^a/x^b)^1/ab . (x^b/x^c)^1/bc . (x^c/x^a)^1/ac=?

Answer 1

$= > (\frac{ {x}^{a} }{ {x}^{b} } {})^{ \frac{1}{ab} } \times ( \frac{ {x}^{b} }{ {x}^{c} } ) {}^{ \frac{1}{bc} } \times ( \frac{ {x}^{c} }{ {x}^{a} } ) {}^{ \frac{1}{ac} }$



By the properties of Exponents.



$= > { ({x}^{a - b}) }^{ \frac{1}{ab} } \times { {(x}^{b - c} )}^{bc} \times { {(x}^{c - a} )}^{ \frac{1}{ac} }$



$= > {x}^{(a - b) \times \frac{1}{ab} } \: \: \times \: \: {x}^{ (b - c) \times \frac{1}{ bc}} \: \: \times \: \: {x}^{(c - a) \times \frac{1}{ac} }$



$= > {x}^{ \frac{a - b}{ab} } \: \: \times \: \: {x}^{ \frac{b - c}{bc} } \: \: \times \: \: {x}^{ \frac{c - a}{ca} }$




$= > x \frac{(a - b)c + (b - c)a + (c - a)b}{abc}$



$= > {x}^{ ac - bc + ab - ac + bc - ab} \\ \\ \\ \\ = > {x}^{ \cancel{ac} - \cancel{ac} - \cancel{ bc} + \cancel{bc} + \cancel{ab} - \cancel{ab}} \\ \\ \\ = > {x}^{0}$




From the properties of Exponents, we know that anything having 0 as its power is equal to 1


= > 1




Answer : 1