Question

The value of $\dfrac{{{x^{a + b}}{x^{b + c}}{x^{c + a}}}}{{{{\left( {{x^a}{x^b}{x^c}} \right)}^2}}}$ is
    (A) ${x^2}$
    (B) ${x^{a + b + c}}$
    (C) ${x^{abc}}$
    (D) ${x^0}$

Answer 1

Hint- In order to solve such type of question, we will try to break the power of x with the help of simple formulas as \[{x^{p + q}} = {x^p}{x^q}\] , then we will proceed further by simplifying them.

Complete step-by-step solution -
Given that $\dfrac{{{x^{a + b}}{x^{b + c}}{x^{c + a}}}}{{{{\left( {{x^a}{x^b}{x^c}} \right)}^2}}}$
We have to break the power of x, and then simplify
As we know \[{x^{p + q}} = {x^p}{x^q}\]
The given equation becomes
$
   \Rightarrow \dfrac{{{x^{a + b}}{x^{b + c}}{x^{c + a}}}}{{{{\left( {{x^a}{x^b}{x^c}} \right)}^2}}} \\
   \Rightarrow \dfrac{{{x^{a + b + b + c + c + a}}}}{{{{\left( {{x^{a + b + c}}} \right)}^2}}} \\
   \Rightarrow \dfrac{{{x^{2a + 2b + 2c}}}}{{{x^{2 \times \left( {a + b + c} \right)}}}} \\
   \Rightarrow \dfrac{{{x^{2a + 2b + 2c}}}}{{{x^{2a + 2b + 2c}}}} \\
   \Rightarrow {x^{\left( {2a + 2b + 2c} \right) - 2a + 2b + 2c}} = {x^0} \\
 $
Hence, the value of $\dfrac{{{x^{a + b}}{x^{b + c}}{x^{c + a}}}}{{{{\left( {{x^a}{x^b}{x^c}} \right)}^2}}}$ is ${x^0}$
So, the correct option is “D”.

Note- The above problem belongs to exponents and can be solved by expanding the exponent terms and cancelling the like terms. The few properties of exponents are : the exponents tell us how many times to use the number in multiplication, A negative exponent means divide because the opposite of multiplying is dividing and a fraction exponent means to take the root.