Question

The numerical value of ${{\left( {{x}^{\dfrac{1}{a-b}}} \right)}^{\dfrac{1}{a-c}}}\times {{\left( {{x}^{\dfrac{1}{b-c}}} \right)}^{\dfrac{1}{b-a}}}\times {{\left( {{x}^{\dfrac{1}{c-a}}} \right)}^{\dfrac{1}{c-b}}}$ is ($a,b,c$ are distinct real numbers)
$A)1$
$B)8$
$C)0$
$D)None$

Answer 1

Hint: To solve the question we need to know the concept of exponents and powers. The formula need to be used here is ${{a}^{p}}\times {{a}^{q}}\times {{a}^{r}}={{a}^{p+q+r}}$ and ${{\left( {{a}^{p}} \right)}^{q}}={{a}^{p\times q}}$. The first step is to simplify the exponent in the expression into the most suitable way. The second step is to find the value of ${{x}^{\text{exponent}}}$ .

Complete step by step answer:
The question ask us to find the value of the expression which is given as${{\left( {{x}^{\dfrac{1}{a-b}}} \right)}^{\dfrac{1}{a-c}}}\times {{\left( {{x}^{\dfrac{1}{b-c}}} \right)}^{\dfrac{1}{b-a}}}\times {{\left( {{x}^{\dfrac{1}{c-a}}} \right)}^{\dfrac{1}{c-b}}}$ . The first step in this process is to simplify the exponents. For that we will have to apply some formulas. Since we know that, when a number has a power $p$ whose power is $q$ then the power gets multiplied. The formula is ${{\left( {{a}^{p}} \right)}^{q}}={{a}^{p\times q}}$. On applying the same in the exponents of the expression given to us, on doing so we get:
 \[= \left( {{x}^{\dfrac{1}{a-b}\times \dfrac{1}{a-c}}} \right)\times \left( {{x}^{\dfrac{1}{b-c}\times \dfrac{1}{b-a}}} \right)\times \left( {{x}^{\dfrac{1}{c-a}\times \dfrac{1}{c-b}}} \right)\]
On further calculation we get:
\[= {{x}^{\dfrac{1}{\left( a-b \right)\left( a-c \right)}}}\times {{x}^{\dfrac{1}{\left( b-c \right)\left( b-a \right)}}}\times {{x}^{\dfrac{1}{\left( c-a \right)\left( c-b \right)}}}\]

Now we will again have to apply the rule of exponent, which says that when the same number with different powers is multiplied then the powers get added. On formulation of the rule we get ${{a}^{p}}\times {{a}^{q}}\times {{a}^{r}}={{a}^{p+q+r}}$. Applying the same in the expression we have, so the expression becomes:
\[= {{x}^{\dfrac{1}{\left( a-b \right)\left( a-c \right)}+\dfrac{1}{\left( b-c \right)\left( b-a \right)}+\dfrac{1}{\left( c-a \right)\left( c-b \right)}}}\]
The next step will be to find the L.C.M of all the denominators in the exponent.
\[= {{x}^{\dfrac{1}{\left( a-b \right)\left( a-c \right)}+\dfrac{1}{-\left( b-c \right)\left( a-b \right)}+\dfrac{1}{\left( a-c \right)\left( b-c \right)}}}\]
\[= {{x}^{\dfrac{(b-c)-(a-c)+(a-b)}{\left( a-b \right)\left( a-c \right)(b-c)}}}\]
On opening the brackets to calculate the numerator of the exponent we get:
\[= {{x}^{\dfrac{b-c-a+c+a-b}{\left( a-b \right)\left( a-c \right)(b-c)}}}\]
We see that all the terms get cancelled as the terms in the numerator turn to zero.
\[= {{x}^{\dfrac{0}{\left( a-b \right)\left( a-c \right)(b-c)}}}\]
\[= {{x}^{0}}\]
Now as we know that, any number having power as zero results to $1$. So \[{{x}^{0}}=1\].
$\therefore $ The numerical value of ${{\left( {{x}^{\dfrac{1}{a-b}}} \right)}^{\dfrac{1}{a-c}}}\times {{\left( {{x}^{\dfrac{1}{b-c}}} \right)}^{\dfrac{1}{b-a}}}\times {{\left( {{x}^{\dfrac{1}{c-a}}} \right)}^{\dfrac{1}{c-b}}}$ is $1$.

So, the correct answer is “Option A”.

Note: Additional to the above rules of exponents used in the question, there are other rules too which would be required in other problems. One of them is the division rule, which says if we need to solve for ${{a}^{p}}\div {{a}^{q}}$ without proper solving, then we could find it using $\dfrac{{{a}^{p}}}{{{a}^{q}}}$ , which in turn results to ${{a}^{p-q}}$ .