Question

If x, y and z are distinct positive numbers such that x + 1/y = y + 1/z = z + 1/x, then the value of xyz is - _______.​

Answer 1

Required Answer:-

Any perfect cubes.

Before Solving:-

A unit fraction of a natural number lies between 0 and 1 and can be 1. We see three numbers are not distinct, but by inspection they cannot equal, because unit fractions are always less than or equal to 1. Thus, we are going to prove the distinctness cannot exist.

Solution:-

Let's assume they are distinct, and assume $x> y> z\geq 1$.

We obtain three inequalities,

$\therefore x+1>x,x\geq y+1,y\geq z+1$ …[1]

Let's take inverses of three numbers,

$1\geq \dfrac{1}{z} >\dfrac{1}{y} > \dfrac{1}{x} >0$ …[2]

Thus, by [1] and [2],

• $x+1>x+\dfrac{1}{y} >x$
• $x\geq y+1\geq y+\dfrac{1}{z}>y$
• $y\geq z+1>z+\dfrac{1}{x}>z$

$\therefore x+\dfrac{1}{y} >y+\dfrac{1}{z} >z+\dfrac{1}{x}$, this is a contradiction. So the assumption is wrong, and hence $x=y=z$. The value of $xyz$ is a perfect cube.

Answer 2

Step-by-step explanation:

perfect cube please mark as best answer and thank me