Prove that ${{\left( {{x}^{a}}/{{x}^{b}} \right)}^{\dfrac{1}{ab}}}{{\left( {{x}^{b}}/{{x}^{c}} \right)}^{\dfrac{1}{bc}}}{{\left( {{x}^{c}}/{{x}^{a}} \right)}^{\dfrac{1}{ca}}}=1$.

Hint: In this question we have been given an expression which has exponential terms in multiplication and we have to prove that the left-hand side of the expression is equal to the right-hand side of the expression. We will solve this question by first considering the left-hand hand side of the expression and using the various properties of exponents, to prove that the left-hand side is equal to the right-hand side.

Complete step by step answer:
We have the expression given to us as:
$\Rightarrow {{\left( {{x}^{a}}/{{x}^{b}} \right)}^{\dfrac{1}{ab}}}{{\left( {{x}^{b}}/{{x}^{c}} \right)}^{\dfrac{1}{bc}}}{{\left( {{x}^{c}}/{{x}^{a}} \right)}^{\dfrac{1}{ca}}}=1$
Consider the left-hand side of the expression, we get:
$\Rightarrow {{\left( {{x}^{a}}/{{x}^{b}} \right)}^{\dfrac{1}{ab}}}{{\left( {{x}^{b}}/{{x}^{c}} \right)}^{\dfrac{1}{bc}}}{{\left( {{x}^{c}}/{{x}^{a}} \right)}^{\dfrac{1}{ca}}}$
Now we know the property of exponents that $\dfrac{{{m}^{p}}}{{{m}^{q}}}={{m}^{p-q}}$ therefore, on substituting, we get:
 $\Rightarrow {{\left( {{x}^{a-b}} \right)}^{\dfrac{1}{ab}}}{{\left( {{x}^{b-c}} \right)}^{\dfrac{1}{bc}}}{{\left( {{x}^{c-a}} \right)}^{\dfrac{1}{ca}}}$
Now we know the property that ${{\left( {{m}^{p}} \right)}^{q}}={{m}^{pq}}$therefore, on substituting, we get:
$\Rightarrow {{\left( x \right)}^{\dfrac{a-b}{ab}}}{{\left( x \right)}^{\dfrac{b-c}{bc}}}{{\left( x \right)}^{\dfrac{c-a}{ca}}}$
$\Rightarrow {{\left( x \right)}^{\dfrac{a-b}{ab}+\dfrac{b-c}{bc}+\dfrac{c-a}{ca}}}$
Now since the bases of all the exponents are same, we can write:
$\Rightarrow {{\left( x \right)}^{\dfrac{a-b}{ab}+\dfrac{b-c}{bc}+\dfrac{c-a}{ca}}}$
On taking the lowest common multiple of the terms, we get:
\[\Rightarrow {{\left( x \right)}^{\dfrac{ac-bc+ab-ac+bc-ab}{abc}}}\]
On adding the terms in the numerator, we get:
\[\Rightarrow {{\left( x \right)}^{\dfrac{0}{abc}}}\]
Since $0$ divided by anything is $0$, we get:
\[\Rightarrow {{\left( x \right)}^{0}}\]
Now we know that anything raised to $0$ is $1$ therefore, we can write:
\[\Rightarrow {{\left( x \right)}^{0}}=1\], which is the right-hand side of the expression, hence proved.

Note: The various properties regarding exponents should be remembered while doing these types of questions. If unable to solve the problem by using rules of exponents, logarithm should be used to simplify the exponential expressions into logarithmic expressions and then the various properties of logarithms should be used to solve the problem. Logarithm is used to simplify a mathematical expression, it converts multiplication to addition, division to subtraction and exponents to multiplication.