Question

Solve the following algebraic expression :
\[{{\left( \dfrac{{{x}^{b}}}{{{x}^{c}}} \right)}^{b+c-a}}{{\left( \dfrac{{{x}^{c}}}{{{x}^{a}}} \right)}^{c+a-b}}{{\left( \dfrac{{{x}^{a}}}{{{x}^{b}}} \right)}^{a+b-c}}\]

Answer 1

Hint: At first, use the identity \[\dfrac{{{x}^{r}}}{{{x}^{s}}}\text{ as }{{x}^{r-s}}\] and then \[{{\left( {{x}^{r}} \right)}^{s}}\] is equal to \[{{x}^{rs}}\] and finally use the identity \[{{x}^{r}}\times {{x}^{s}}={{x}^{r+s}}\] to solve the question.

Complete step-by-step solution -
In the question, we are given an expression \[{{\left( \dfrac{{{x}^{b}}}{{{x}^{c}}} \right)}^{b+c-a}}{{\left( \dfrac{{{x}^{c}}}{{{x}^{a}}} \right)}^{c+a-b}}{{\left( \dfrac{{{x}^{a}}}{{{x}^{b}}} \right)}^{a+b-c}}\] and we have to solve it. Now, before proceeding, we will use an identity of the exponents which is \[{{x}^{a-b}}=\dfrac{{{x}^{a}}}{{{x}^{b}}}\] where x is called the base and a, b are exponents, so we are given the expression \[{{\left( \dfrac{{{x}^{b}}}{{{x}^{c}}} \right)}^{b+c-a}}\times {{\left( \dfrac{{{x}^{c}}}{{{x}^{a}}} \right)}^{c+a-b}}\times {{\left( \dfrac{{{x}^{a}}}{{{x}^{b}}} \right)}^{a+b-c}}\] and we can write the given expression as
\[{{\left( {{x}^{\left( b-c \right)}} \right)}^{b+c-a}}\times {{\left( {{x}^{\left( c-a \right)}} \right)}^{c+a-b}}\times {{\left( {{x}^{\left( a-b \right)}} \right)}^{a+b-c}}\]
We can further simplify and write it as,
\[{{x}^{\left( b-c \right)\left( b+c-a \right)}}.{{x}^{\left( c-a \right)\left( c+a-b \right)}}.{{x}^{\left( a-b \right)\left( a+b-c \right)}}\]
This expression we wrote by using the formula, \[{{\left( {{x}^{c}} \right)}^{d}}={{x}^{cd}}.\]
Now, the expression stands is
\[{{x}^{\left( b-c \right)\left( b+c-a \right)}}\times {{x}^{\left( c-a \right)\left( c+a-b \right)}}\times {{x}^{\left( a-b \right)\left( a+b-c \right)}}\]
Now, we will first find the products of the expressions (b – c) (b + c – a), (c – a) (c + a – b) and (a – b) (a + b – c). So, the product of (b – c) (b + c – a) is
(b – c) (b + c) – a (b – c)
\[\Rightarrow {{b}^{2}}-{{c}^{2}}-ab+ac\]
The product of (c – a) (c + a – b) is
(c – a) (c + a) – b (c – a)
\[\Rightarrow {{c}^{2}}-{{a}^{2}}-bc+ab\]
And the product of (a – b) (a + b – c) is
(a – b) (a + b) – c (a – b)
\[\Rightarrow {{a}^{2}}-{{b}^{2}}-ca+bc\]
So, after finding the products, we can write the expression as,
\[{{x}^{{{b}^{2}}-{{c}^{2}}-ab+ac}}\times{{x}^{{{c}^{2}}-{{a}^{2}}-bc+ab}}\times {{x}^{{{a}^{2}}-{{b}^{2}}-ca+bc}}\]
Now, we will apply the identity,
\[{{x}^{r}}\times {{x}^{s}}={{x}^{r+s}}\]
So, by using this, we can write the expression as,
 \[{{x}^{{{b}^{2}}-{{c}^{2}}-ab+ac+{{c}^{2}}-{{a}^{2}}-bc+ab+{{a}^{2}}-{{b}^{2}}-ca+bc}}\]
Now, on simplifying the powers of x, we get, \[{{x}^{0}}\] and we know that any number to the power 0 is 1.
Hence, the value is 1.

Note: There is also another way to solve by dealing with the numerator and denominator differently and then use the identity \[\dfrac{{{x}^{r}}}{{{x}^{s}}}={{x}^{r-s}}\] to solve to get the answer.