Question

If ${X^X} \cdot {Y^Y} \cdot {Z^Z} = {X^Y} \cdot {Y^Z} \cdot {Z^X} = {X^Z} \cdot {Y^X} \cdot {Z^Y}$ such that $X$, $Y$ and $Z$ are positive integers greater than $1$, then which of the following can’t be true for any of the positive value of $X$, $Y$ and $Z$?
(A) $XYZ = 27$
(B) $XYZ = 1728$
(C) $X + Y + {\rm Z} = 32$
(D) $X + Y + Z = 12$

Answer 1

Hint: The variables \[X\], $Y$ and $Z$ satisfy the relation ${X^X} \cdot {Y^Y} \cdot {Z^Z} = {X^Y} \cdot {Y^Z} \cdot {Z^X} = {X^Z} \cdot {Y^X} \cdot {Z^Y}$. So, our main target is to find the values of variables or the relation between these variables. Then one by one check the given options.

Complete step by step solution:
The given relation between the variables \[X\], $Y$ and $Z$ is ${X^X} \cdot {Y^Y} \cdot {Z^Z} = {X^Y} \cdot {Y^Z} \cdot {Z^X} = {X^Z} \cdot {Y^X} \cdot {Z^Y}$. It is also given that we have to check only for the positive values of the variables $X$, $Y$ and $Z$.
We can say ${X^X} \cdot {Y^Y} \cdot {Z^Z} = {X^Y} \cdot {Y^Z} \cdot {Z^X}$
Now, we can write this equation as $\dfrac{{{X^X} \cdot {Y^Y} \cdot {Z^Z}}}{{{X^Y} \cdot {Y^Z} \cdot {Z^X}}} = 1$.
We know an algebraic identity $\dfrac{{{a^m}}}{{{a^n}}} = {a^{m - n}}$. Using the given identity we can write the above equation as
$
   \Rightarrow \dfrac{{{X^X} \cdot {Y^Y} \cdot {Z^Z}}}{{{X^Y} \cdot {Y^Z} \cdot {Z^X}}} = 1 \\
   \Rightarrow {X^{X - Y}} \cdot {Y^{Y - Z}} \cdot {Z^{Z - X}} = 1
 $
On the right side of the above equation is $1$, meaning the power of the variables \[X\], $Y$ and $Z$ are zero.
This implies $\left( {X - Y} \right) = 0$, $\left( {Y - Z} \right) = 0$ and $\left( {Z - X} \right) = 0$.
This means $X = Y$ , $Y = Z$ and $Z = X$.
Therefore, the variables $X$, $Y$ and $Z$ are equal. That is $X = Y = Z$.
Now, we have to check each option one by one.
Option (A)
$XYZ = 27$
Since, the variables $X$, $Y$ and $Z$ are equal. So, we can write $XYZ = 27$ as $X \times X \times X = 27$.
$
   \Rightarrow {X^3} = 27 \\
  \therefore X = \sqrt[3]{{27}} = 3
 $
We get $X = Y = Z = 3$, which is a positive integer. So, the option (A) is true.
Option (B)
$XYZ = 1728$
Since, the variables $X$, $Y$ and $Z$ are equal. So, we can write $XYZ = 1728$ as $X \times X \times X = 1728$.
$
   \Rightarrow {X^3} = 1728 \\
  \therefore X = \sqrt[3]{{1728}} = 12
 $
We get $X = Y = Z = 12$, which is a positive integer. So, the option (B) is true.
Option (C)
$X + Y + {\rm Z} = 32$
Since, the variables $X$, $Y$ and $Z$ are equal. So, we can write $X + Y + Z = 32$ as $X + X + X = 32$.
$
   \Rightarrow 3X = 32 \\
  \therefore X = \dfrac{{32}}{3}
 $
We get $X = Y = Z = \dfrac{{32}}{3}$, which is not a positive integer. So, the option (C) is not true.
Option (D)
$X + Y + Z = 12$
 Since, the variables $X$, $Y$ and $Z$ are equal. So, we can write $X + Y + Z = 12$ as $X + X + X = 12$.
$
   \Rightarrow 3X = 12 \\
  \therefore X = \dfrac{{12}}{3} = 4
 $
We get $X = Y = Z = 4$, which is a positive integer. So, the option (D) is true.

Hence, option (C) is correct for this question.

Note: The given relation is not possible if any on the variable $X$, $Y$ and $Z$ is equal to zero because $\dfrac{a}{0}$ is not defined. This relation is not true for any negative value of the variables because the value of ${X^X}{Y^Y}{Z^Z}$ comes to be imaginary.