Question

Prove that :
\[{\left( {\dfrac{{{x^a}}}{{{x^b}}}} \right)^c}\times {\left( {\dfrac{{{x^b}}}{{{x^c}}}} \right)^a}\times {\left( {\dfrac{{{x^c}}}{{{x^a}}}} \right)^b}\]\[ = 1\]

Answer 1

Hint:
To approach this question the things we need to keep in mind is that x being the common base so the powers in the division get subtracted and the powers involved in the multiplication with common base x are added to each other. As the different powers with the common base are subtracted if the variable with the same base and different powers are involved in division and the different powers sharing the same base are added if the variable with same base and different powers are involved in multiplication. so, in this question the powers associated with base x with in division that is \[{\left( {\dfrac{{{x^a}}}{{{x^b}}}} \right)^c}'{\left( {\dfrac{{{x^b}}}{{{x^c}}}} \right)^a}'{\left( {\dfrac{{{x^c}}}{{{x^a}}}} \right)^b}\] are first subtracted as power in the denominator gets subtracted from numerator as far as for the concern of the powers outside the brackets that are involved in the multiplication of this expression that is \[{\left( {\dfrac{{{x^a}}}{{{x^b}}}} \right)^c}\times {\left( {\dfrac{{{x^b}}}{{{x^c}}}} \right)^a}\times {\left( {\dfrac{{{x^c}}}{{{x^a}}}} \right)^b}\] are added to each other and the resultant is calculated then and further solved to get the desired answer.

Complete step by step solution:
Here taking Left Hand Side that is –
\[{\left( {\dfrac{{{x^a}}}{{{x^b}}}} \right)^c}\times {\left( {\dfrac{{{x^b}}}{{{x^c}}}} \right)^a}\times {\left( {\dfrac{{{x^c}}}{{{x^a}}}} \right)^b}\]
Firstly solving the fraction inside the brackets as the different powers with the common base are subtracted if the variable with same base and different powers are involved in division that is \[\dfrac{{{t^u}}}{{{t^v}}} = {t^{u - v}}\]
Applying it we would get –
\[{\left( {{x^{a - b}}} \right)^c}\times {\left( {{x^{b - c}}} \right)^a}\times {\left( {{x^{c - a}}} \right)^b}\]
Now opening the brackets and multiplying the respective powers we get -:
 \[{\left( {{x^{ac - bc}}} \right)^{}}\times \left( {{x^{ab - ac}}} \right)\times \left( {{x^{bc - ab}}} \right)\]
Now again opening the brackets and further solving by applying the rule that the different powers sharing same base are added if the variable with same base and different powers are involved in multiplication that is \[{t^u}\times {t^v} = {t^{u + v}}\]
So applying it in the equation\[{\left( {{x^{ac - bc}}} \right)^{}}\times \left( {{x^{ab - ac}}} \right)\times \left( {{x^{bc - ab}}} \right)\] we get –
\[\left( {{x^{ac - bc + ab - ac + bc - ab}}} \right)\]
Solving the powers further we get-
\[{x^0}\]\[ = 1\]
As something raised to power 0 is 1 that is –
\[{t^0} = 1\]
So resulting answer is 1
Hence left hand side is equal to right hand side
Hence proved \[{\left( {\dfrac{{{x^a}}}{{{x^b}}}} \right)^c}\times {\left( {\dfrac{{{x^b}}}{{{x^c}}}} \right)^a}\times {\left( {\dfrac{{{x^c}}}{{{x^a}}}} \right)^b}\]\[ = 1\]

Note:
Another simpler that can be used to solve such kind of questions is that common base involving all three variables of powers in 3 different ways that is \[{\left( {\dfrac{{{t^a}}}{{{t^b}}}} \right)^c}'{\left( {\dfrac{{{t^b}}}{{{t^c}}}} \right)^a}'{\left( {\dfrac{{{t^c}}}{{{t^a}}}} \right)^b}\]generally have results as 1.