Question

If x is a positive real number and the exponents are rational numbers then show that:
\[{{\left[ \dfrac{{{x}^{a}}}{{{x}^{b}}} \right]}^{a+b-c}}{{\left[ \dfrac{{{x}^{b}}}{{{x}^{c}}} \right]}^{b+c-a}}{{\left[ \dfrac{{{x}^{c}}}{{{x}^{a}}} \right]}^{c+a-b}}=1\]

Answer 1

Hint: We solve this problem by using some of the formulas of exponents. We take the LHS of the given equation then we prove that the LHS is equal to RHS.
We have some formulas that are
(1) \[{{\left( \dfrac{p}{q} \right)}^{r}}=\dfrac{{{p}^{r}}}{{{q}^{r}}}\]
(2) \[\dfrac{{{p}^{q}}}{{{p}^{r}}}={{p}^{q-r}}\]
(3) \[{{\left( {{p}^{q}} \right)}^{r}}={{p}^{q\times r}}\]
(4) \[{{p}^{q}}\times {{p}^{r}}={{p}^{q+r}}\]
By using the above formulas in the LHS of the given equation we can prove the required result.

Complete step-by-step solution
We know that the formula of exponents that is
\[\dfrac{{{p}^{q}}}{{{p}^{r}}}={{p}^{q-r}}\]
By using this formula to above equation we get
\[\begin{align}
  & \Rightarrow LHS={{\left( {{x}^{a}} \right)}^{a+b-c-c-a+b}}\times {{\left( {{x}^{b}} \right)}^{b+c-a-a-b+c}}\times {{\left( {{x}^{c}} \right)}^{c+a-b-b-c+a}} \\
 & \Rightarrow LHS={{\left( {{x}^{a}} \right)}^{2b-2c}}\times {{\left( {{x}^{b}} \right)}^{2c-2a}}\times {{\left( {{x}^{c}} \right)}^{2a-2b}} \\
\end{align}\]
We know that the formula of exponents that is
\[{{\left( {{p}^{q}} \right)}^{r}}={{p}^{q\times r}}\]
By using this formula to above equation we get
\[\Rightarrow LHS={{x}^{2ab-2ac}}\times {{x}^{2bc-2ab}}\times {{x}^{2ac-2bc}}\]
We know that the formula of exponents that is
\[{{p}^{q}}\times {{p}^{r}}={{p}^{q+r}}\]
By using this formula to above equation we get
\[\begin{align}
  & \Rightarrow LHS={{x}^{2ab-2ac+2bc-2ab+2ac-2bc}} \\
 & \Rightarrow LHS={{x}^{0}}=1 \\
 & \Rightarrow LHS=RHS \\
\end{align}\]
Here, we can see that the LHS and RHS of the given equation are equal.
So, we can conclude that the given equation has been proved that is
\[\therefore {{\left[ \dfrac{{{x}^{a}}}{{{x}^{b}}} \right]}^{a+b-c}}{{\left[ \dfrac{{{x}^{b}}}{{{x}^{c}}} \right]}^{b+c-a}}{{\left[ \dfrac{{{x}^{c}}}{{{x}^{a}}} \right]}^{c+a-b}}=1\]

Note: Students may do mistake by directly adding the terms of the exponents that is we have the LHS of the given equation as
\[\Rightarrow LHS={{\left[ \dfrac{{{x}^{a}}}{{{x}^{b}}} \right]}^{a+b-c}}{{\left[ \dfrac{{{x}^{b}}}{{{x}^{c}}} \right]}^{b+c-a}}{{\left[ \dfrac{{{x}^{c}}}{{{x}^{a}}} \right]}^{c+a-b}}\]
We know that the formulas of exponents that are
(1) \[{{\left( \dfrac{p}{q} \right)}^{r}}=\dfrac{{{p}^{r}}}{{{q}^{r}}}\]
(2) \[\dfrac{{{p}^{q}}}{{{p}^{r}}}={{p}^{q-r}}\]
(3) \[{{\left( {{p}^{q}} \right)}^{r}}={{p}^{q\times r}}\]
(4) \[{{p}^{q}}\times {{p}^{r}}={{p}^{q+r}}\]
They apply all at once without any rearrangements or regrouping as follows
\[\begin{align}
  & \Rightarrow LHS={{x}^{{{a}^{2}}+ab-ac-ab-{{b}^{2}}+bc+{{b}^{2}}+bc-ca-cb-{{c}^{2}}+ac+{{c}^{2}}+ac-cb-ac-{{a}^{2}}+ab}} \\
 & \Rightarrow LHS={{x}^{0}}=1 \\
\end{align}\]
Here, students may miss some of the terms during this multiplication in this process. So, we need to rearrange or regroup so as to avoid the confusion.