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**Hint:**We will first of all mention some formulas and identities which are going to be used for the solution of the above given question and then use them to prove the required..

**Complete answer:**

We know that we have some formulas given by the following expressions:-

$ \Rightarrow \dfrac{{{u^x}}}{{{u^y}}} = {u^{x - y}}$ ……………(1)

$ \Rightarrow {\left( {{a^b}} \right)^c} = {a^{bc}}$ …………..(2)

$ \Rightarrow \left( {{a^b}} \right) \times \left( {{a^c}} \right) = {a^{b + c}}$ …………..(3)

Now, we are given the left hand side as the expression: ${\left( {\dfrac{{{x^a}}}{{{x^b}}}} \right)^c} \times {\left( {\dfrac{{{x^b}}}{{{x^c}}}} \right)^a} \times {\left( {\dfrac{{{x^c}}}{{{x^a}}}} \right)^b}$

Let us assume that $u = {\left( {\dfrac{{{x^a}}}{{{x^b}}}} \right)^c},v = {\left( {\dfrac{{{x^b}}}{{{x^c}}}} \right)^a}$ and $w = {\left( {\dfrac{{{x^c}}}{{{x^a}}}} \right)^b}$

Now, we will first of all consider u only.

We have $u = {\left( {\dfrac{{{x^a}}}{{{x^b}}}} \right)^c}$

Using the formula in equation (1), we can write the above expression as following:-

$ \Rightarrow u = {\left( {{x^{a - b}}} \right)^c}$

Using the formula in equation (2), we can write the above expression as following:-

$ \Rightarrow u = {x^{c\left( {a - b} \right)}}$

Simplifying the powers on top, we will get:-

$ \Rightarrow u = {x^{ca - cb}}$ ………………..(4)

Now, let us consider v only.

We have $v = {\left( {\dfrac{{{x^b}}}{{{x^c}}}} \right)^a}$

Using the formula in equation (1), we can write the above expression as following:-

$ \Rightarrow v = {\left( {{x^{b - c}}} \right)^a}$

Using the formula in equation (2), we can write the above expression as following:-

$ \Rightarrow v = {x^{a\left( {b - c} \right)}}$

Simplifying the powers on top, we will get:-

$ \Rightarrow v = {x^{ab - ac}}$ ………………..(5)

Now, let us consider w only.

We have $w = {\left( {\dfrac{{{x^c}}}{{{x^a}}}} \right)^b}$

Using the formula in equation (1), we can write the above expression as following:-

$ \Rightarrow w = {\left( {{x^{c - a}}} \right)^b}$

Using the formula in equation (2), we can write the above expression as following:-

$ \Rightarrow w = {x^{b\left( {c - a} \right)}}$

Simplifying the powers on top, we will get:-

$ \Rightarrow w = {x^{bc - ba}}$ ………………..(6)

Now, our left hand side was u.v.w. So, if we use the equations (4), (5) and (6), we will then obtain the following expression:-

$ \Rightarrow $L. H. S. = $u \times v \times w = {x^{ca - cb}} \times {x^{ab - ac}} \times {x^{bc - ba}}$

Now, we will use the equation number (3) in the above derived expression to get the following expression:-

$ \Rightarrow $L. H. S. = ${x^{ca - cb + ab - ac + bc - ba}}$

Simplifying the power above, we will then obtain:-

$ \Rightarrow $L. H. S. = ${x^0}$ = 1 = R. H. S.

**Note:**

The students must note that any real number when raised to the power 0 will always be equal to 1, no matter what. Therefore, in the last step we had ${x^0} = 1$.

The students must commit to memory the following formulas:-

$ \Rightarrow \dfrac{{{u^x}}}{{{u^y}}} = {u^{x - y}}$

$ \Rightarrow {\left( {{a^b}} \right)^c} = {a^{bc}}$

$ \Rightarrow \left( {{a^b}} \right) \times \left( {{a^c}} \right) = {a^{b + c}}$

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