The T(h)rees

Algebra Level 3

Consider three real numbers $(x,y,z)$ such that

$\dfrac{x^3 - 9}{y} = \dfrac{y^3 - 9}{z} = \dfrac{z^3 - 9}{x} = 6$

Evaluate $x^y + y^z + z^x$ .

The answer is 81.

4 solutions

Guilherme Dela Corte
Apr 3, 2014

$x^3 - 9 = 6y, \; y^3 - 9 = 6z, \; z^3 - 9 = 6x$

$\color{#D61F06}{x \geq y \geq z}$

$x^3 \geq y^3 \geq z^3 \Leftrightarrow x^3 - 9 \geq y^3 - 9 \geq z^3 - 9$

$\Leftrightarrow \color{#3D99F6}{6y \geq 6z \geq 6x} \Rightarrow \color{#3D99F6}{z \geq x} \; \wedge \; \color{#D61F06}{x \geq z} \Rightarrow x=y=z$

$x^3 - 9 = 6x \Leftrightarrow (x-3)(x^2 +3x +3)=0 \Rightarrow x \in \mathbb{R}, \; x = 3$

$\boxed{x^y + y^z + z^x = 81}$

Nice proof to your problem. I have a short querry. How do you show

$x \geq y \geq z$ will yield $x^3 \geq y^3 \geq z^3$ ? Will this always work? Thank you!

Datu Oen - 7 years, 2 months ago

For real numbers that will always work.

For positive numbers : multiplying both sides by a positive number will still be true. Thus, $x^2\geq xy$ and $xy \geq y^2$ . Using these two new inequalities, we have $x^2 \geq y^2$ . Doing the same process again, we have $x^3 \geq y^2x$ and $x^2y \geq y^3$ .

For negative numbers : multiplying both sides by a positive number will still be true. Thus, $-x^2\geq -xy$ and $-xy \geq -y^2$ . Using these two new inequalities, we have $-x^2 \geq -y^2$ . Doing the same process again, we have $x^3 \geq y^2x$ and $x^2y \geq y^3$ .

Notice: for both cases we must have that $x^2y \geq y^2x$ , proved true by $xy \cdot x \geq xy \cdot y \Leftrightarrow x \geq y$ . Thus $x^3 \geq y^3$ .

Guilherme Dela Corte - 7 years, 2 months ago

Thank you.

I think your proof on the above problem is safe with the equality under the inequality (greater than or equal to ). This is a strong statement which forces $x, y, z$ to be equal.

Datu Oen - 7 years, 2 months ago

Nice soultions

Kahsay Merkeb - 7 years, 2 months ago

Datu Oen
Apr 3, 2014

Suppose that $x = y = z$ , then $\displaystyle\frac{x^3 - 9}{x} = 6$ would lead to $x^3 - 9 = 6x \hspace{1cm} \mbox{that is} \hspace{1cm} x^3 - 6x - 9 = 0$ .

$x^3 - 6x - 9 = (x - 3) (x^2 + 3x + 3) = 0$ Since $x$ is a real number, then $x = 3$ . Thus, $x^y + y^ z + z^x = 3 \times x^x = 3 \times (3)^3 = 81$

How did you prove $x = y = z$ ?

Guilherme Dela Corte - 7 years, 2 months ago

Maybe let us try (semi) proof using method of contradiction.

Suppose $x = y$ and $x \neq z$ . Then $\displaystyle\frac{x^3-9}{y}$ and $\displaystyle\frac{y^3-9}{z}$ becomes $\displaystyle\frac{x^3-9}{x}$ and $\displaystyle\frac{x^3-9}{z}$ .

Since both expressions are equal to 6, we can equate them. That is

$\displaystyle\frac{x^3-9}{x}=\displaystyle\frac{x^3-9}{z}$ .

Since the fractions are equal and their numerators are equal, we are left with $x = z$ which is a contradiction.

Datu Oen - 7 years, 2 months ago

But what if $x \neq y$ , $y \neq z$ and $z \neq x$ ?

Calvin Lin Staff - 7 years, 2 months ago

@Calvin Lin – Let me try and think of that again.

Datu Oen - 7 years, 2 months ago

Chew-Seong Cheong
Jan 23, 2015

Due to the symmetry of the equations, we can assume that $x=y=z$ . Therefore,

$\dfrac {x^3-9}{y} = 6\quad \Rightarrow \dfrac {x^3-9}{x} = 6 \quad \Rightarrow x^3 - 6x - 9 =0 \quad \Rightarrow x = 3 = y = z$

$\Rightarrow x^y + y^z + z^x = 3^4 = \boxed{81}$

"We can assume that $x=y=z$ "? Please note that that is not always true.

Guilherme Dela Corte - 6 years, 4 months ago

Finn Hulse
Apr 3, 2014

Notice that if $x$ , $y$ , and $z$ are 3, then all equations are satisfied. Another way to approach this is by realizing that all the equations are the same with the same result and thus all variables are equal. Plugging that into what it wants us to find, $\boxed{81}$ is obtained.

It is not necessarily true that if an expression is true cyclically, then the values must be equal. For example, if we're solving

$\frac{ x^2 - 9 } { yz} = \frac{ y^2 - 9} {zx} = \frac{ z^2 - 9}{xy } = 6$

Then, this equation is satisfied by any set of roots of $\alpha^3 - 9 \alpha - k = 0$ and clearly the roots need not all be equal. (Do you see why this is true?)

You often need another 'size' argument, like what Guilherme did in his solution, before you can conclude that they are equal.

For more examples, look at my recent spate of questions about symmetric / cyclic expressions in which the 'solution set' doesn't consists of equal terms.

Calvin Lin Staff - 7 years, 2 months ago

That is the case, so I guess I intended to use that method as a possible case, and thus used it to solve that. I was thinking about that other problem you posted similar to this. Do you have any tips on how to manipulate these kind of large algebraic inequality things? They ALWAYS appear on IMO problems and I can never find a good way to solve/factor them.

Finn Hulse - 7 years, 2 months ago

Although it can be a bit tedious, if the solution to a cyclic set of polynomial expressions doesn't jump out at me after some consideration, I tend resort to Gröbner bases as a sure method of solving polynomial systems. If you compute the Gröbner basis of $\left\{x^2-6yz-9, -6xz+y^2-9, -6xy+z^2-9\right\}$ with respect to the lex order $x > y > z$ , then you get $9245z^6-80109z^4-154305z^2+35721 = (5z^2+9)(43z^2-441)(43z^2-9) = 0$ . Thus, in $\mathbb{R}$ , $z = \pm \frac{3}{\sqrt{43}}$ or $z = \pm \frac{21}{\sqrt{43}}$ . The solution values of $x$ and $y$ follow.

Michael Lee - 7 years, 2 months ago

@Michael Lee – Cool. It's funny, I was just reading about that technique but that was after I posted my solution.

Finn Hulse - 7 years, 1 month ago

The T(h)rees

The answer is 81.

4 solutions

0 pending reports