Question

Prove that:
 $ {\left( {\dfrac{{{x^a}}}{{{x^b}}}} \right)^{a + b - c}} \times {\left( {\dfrac{{{x^b}}}{{{x^c}}}} \right)^{b + c - a}} \times {\left( {\dfrac{{{x^c}}}{{{x^a}}}} \right)^{c + a - b}} = 1 $

Answer 1

Hint: Here, we can see that the base of each term is the same. Therefore, to solve this problem, we will use three important rules of exponents. The first rule states that when we need to multiply the numbers having the same base, we should add the exponents. The second rule states that when we need to divide the numbers having the same base, we should subtract the exponents. The third rule states that powers are multiplied when the numbers are raised by another number.
Formulas used:
When we need to multiply the numbers having the same base, we should add the exponents. $ {p^m} \times {p^n} = {p^{\left( {m + n} \right)}} $
When we need to divide the numbers having the same base, we should subtract the exponents.
 $ \dfrac{{{p^m}}}{{{p^n}}} = {p^{\left( {m - n} \right)}} $
 Powers are multiplied when the numbers are raised by another number.
 $ {\left( {{p^m}} \right)^n} = {p^{\left( {m \times n} \right)}} $

Complete step-by-step answer:
We have
 $ L.H.S. = {\left( {\dfrac{{{x^a}}}{{{x^b}}}} \right)^{a + b - c}} \times {\left( {\dfrac{{{x^b}}}{{{x^c}}}} \right)^{b + c - a}} \times {\left( {\dfrac{{{x^c}}}{{{x^a}}}} \right)^{c + a - b}} $
We can rewrite this as
 \[ \Rightarrow L.H.S. = \dfrac{{{{\left( {{x^a}} \right)}^{a + b - c}}}}{{{{\left( {{x^b}} \right)}^{a + b - c}}}} \times \dfrac{{{{\left( {{x^b}} \right)}^{b + c - a}}}}{{{{\left( {{x^c}} \right)}^{b + c - a}}}} \times \dfrac{{{{\left( {{x^c}} \right)}^{c + a - b}}}}{{{{\left( {{x^a}} \right)}^{c + a - b}}}}\]
Here, we will apply the third rule which states that powers are multiplied when the numbers are raised by another number: $ {\left( {{p^m}} \right)^n} = {p^{\left( {m \times n} \right)}} $
 \[ \Rightarrow L.H.S. = \dfrac{{{x^{{a^2} + ab - ac}}}}{{{x^{ab + {b^2} - bc}}}} \times \dfrac{{{x^{{b^2} + bc - ab}}}}{{{x^{bc + {c^2} - ac}}}} \times \dfrac{{{x^{{c^2} + ac - bc}}}}{{{x^{ac + {a^2} - ab}}}}\]
Now we will apply the second rule which states that when we need to divide the numbers having the same base, we should subtract the exponents: $ \dfrac{{{p^m}}}{{{p^n}}} = {p^{\left( {m - n} \right)}} $
 \[
   \Rightarrow L.H.S. = {x^{{a^2} + ab - ac - ab - {b^2} + bc}} \times {x^{{b^2} + bc - ab - bc - {c^2} + ac}} \times {x^{{c^2} + ac - bc - ac - {a^2} + ab}} \\
   \Rightarrow L.H.S. = {x^{{a^2} - ac - {b^2} + bc}} \times {x^{{b^2} - ab - {c^2} + ac}} \times {x^{{c^2} - bc - {a^2} + ab}} \;
 \]
We will now apply the first rule which states that when we need to multiply the numbers having the same base, we should add the exponents: $ {p^m} \times {p^n} = {p^{\left( {m + n} \right)}} $
 \[
   \Rightarrow L.H.S. = {x^{{a^2} - ac - {b^2} + bc + {b^2} - ab - {c^2} + ac + {c^2} - bc - {a^2} + ab}} \\
   \Rightarrow L.H.S. = {x^0} \\
   \Rightarrow L.H.S. = 1 \\
   \Rightarrow L.H.S. = R.H.S \;
 \]
Hence, it is proved that $ {\left( {\dfrac{{{x^a}}}{{{x^b}}}} \right)^{a + b - c}} \times {\left( {\dfrac{{{x^b}}}{{{x^c}}}} \right)^{b + c - a}} \times {\left( {\dfrac{{{x^c}}}{{{x^a}}}} \right)^{c + a - b}} = 1 $

Note: In this problem, we have proved the given equation by using three rules of exponents in the following order:
First, we have multiplied the powers of base $ x $ .
Second, we have subtracted the powers of base $ x $
And finally, we have added the powers of base $ x $ and proved the given equation.