Question

Simplify the following
\[{{\left( \dfrac{{{x}^{b}}}{{{x}^{c}}} \right)}^{a}}\times {{\left( \dfrac{{{x}^{c}}}{{{x}^{a}}} \right)}^{b}}\times {{\left( \dfrac{{{x}^{a}}}{{{x}^{b}}} \right)}^{c}}\].

Answer 1

Hint: For such a type of question we had to simplify these expressions into a simpler form or in a shorter form. We can do this if we know the basic identities to solve the algebraic expressions, and talking about this particular question, we need to know the algebraic identities of exponent one.

Complete step by step solution:
Some identities which we should know to solve these type of questions are:
\[\begin{align}
  & 1).\text{ }{{a}^{0}}=1 \\
 & 2).\text{ }{{a}^{m}}{{a}^{n}}={{a}^{m+n}} \\
 & 3).\text{ }{{a}^{1}}=a \\
 & 4).\text{ }{{1}^{n}}=1 \\
 & 5).\text{ }{{\left( ab \right)}^{n}}={{a}^{n}}{{b}^{n}} \\
 & 6).\text{ }{{a}^{-1}}=\dfrac{1}{a} \\
 & 7).\text{ }{{\left( \dfrac{a}{b} \right)}^{n}}=\dfrac{{{a}^{n}}}{{{b}^{n}}} \\
 & 8).\text{ }{{\left( \dfrac{{{a}^{x}}}{{{b}^{y}}} \right)}^{n}}=\dfrac{{{a}^{xn}}}{{{b}^{yn}}} \\
 & 9).\text{ }{{a}^{x}}\times \dfrac{{{b}^{x}}}{{{a}^{x}}}={{b}^{x}} \\
\end{align}\]
In these properties a, b are constants whereas m, n, x and y can be variable. Here in this equation, a ${{a}^{x}}$ present in the numerator is canceled by the ${{a}^{x}}$present in the denominator.
Now to solve the question, first we will try to open the brackets of the given expression that is \[{{\left( \dfrac{{{x}^{b}}}{{{x}^{c}}} \right)}^{a}}\times {{\left( \dfrac{{{x}^{c}}}{{{x}^{a}}} \right)}^{b}}\times {{\left( \dfrac{{{x}^{a}}}{{{x}^{b}}} \right)}^{c}}\] by using the identity \[{{\left( \dfrac{{{a}^{x}}}{{{b}^{y}}} \right)}^{n}}=\dfrac{{{a}^{xn}}}{{{b}^{yn}}}\]
\[\begin{align}
  & \Rightarrow {{\left( \dfrac{{{x}^{b}}}{{{x}^{c}}} \right)}^{a}}\times {{\left( \dfrac{{{x}^{c}}}{{{x}^{a}}} \right)}^{b}}\times {{\left( \dfrac{{{x}^{a}}}{{{x}^{b}}} \right)}^{c}} \\
 & \Rightarrow \dfrac{{{x}^{ab}}}{{{x}^{ac}}}\times \dfrac{{{x}^{bc}}}{{{x}^{ab}}}\times \dfrac{{{x}^{ac}}}{{{x}^{bc}}} \\
 & \\
\end{align}\]
By using the property \[{{a}^{x}}\times \dfrac{{{b}^{x}}}{{{a}^{x}}}={{b}^{x}}\] from the identities given above, we can see that ${{x}^{ab}}$present in left most of the above expression is also present in the middle term which is in multiplication in the denominator, so by using the equation ‘a’ we can cancel ${{x}^{ab}}$present in left most in the numerator and middle term denominator.
Similarly, is with the ${{x}^{bc}}$which is present in the middle term and rightmost term which is in multiplication, in the numerator and denominator respectively, so by using the equation ‘a’ we can also cancel ${{x}^{bc}}$present in rightmost denominator and middle term numerator.
Same is with the ${{x}^{ac}}$, we can see that ${{x}^{ac}}$present in right most of above expression is also present in the left-most term which is in multiplication in the denominator, so by using the equation ‘a’ we can cancel ${{x}^{ac}}$present in left most in the denominator and rightmost term in numerator.
So all three terms will get canceled by each other and we will get 1.
Hence the answer is 1.

Note: In order to use equation ‘a’ please go through it once, as it says that when two terms are in multiplication and if one term has a value similar to another term in the numerator and denominator respectively or vice versa then only we can directly cancel these values.