1. ## Simultaneous equations

Buenas noches. I have a question here that I have answered, so I hope that the forum could rate my solution, detect errors and provide alternatives. Thanks.

Question: If $\displaystyle x$ and $\displaystyle y$ are distinct real numbers such that$\displaystyle x^3=1-y$ and $\displaystyle y^3=1-x,$ find all possible values of $\displaystyle xy$ .

Solution: $\displaystyle x^3=1-y$
$\displaystyle y^3=1-x,$

$\displaystyle x^3-y^3=x-y$
(Dividing by $\displaystyle (x-y)$ ): $\displaystyle x^2+xy+y^2=1$
$\displaystyle (x+y)^2-xy=1 \Rightarrow(x+y)^2-1^2=xy$
$\displaystyle (x+y+1)(x+y-1)=xy$
Case 1: $\displaystyle x+y+1=x$, $\displaystyle x+y-1=y$,
$\displaystyle y=-1,x=1$
Case 2: $\displaystyle x+y+1=y$, $\displaystyle x+y-1=x$,
$\displaystyle x=-1,y=1$

Only possible value of $\displaystyle xy$ is $\displaystyle -1$

First post with latex.

2. Originally Posted by I-Think
Buenas noches. I have a question here that I have answered, so I hope that the forum could rate my solution, detect errors and provide alternatives. Thanks.

Question: If $\displaystyle x$ and $\displaystyle y$ are distinct real numbers such that$\displaystyle x^3=1-y$ and $\displaystyle y^3=1-x,$ find all possible values of $\displaystyle xy$ .

Solution: $\displaystyle x^3=1-y$
$\displaystyle y^3=1-x,$

$\displaystyle x^3-y^3=x-y$
(Dividing by $\displaystyle (x-y)$ ): $\displaystyle x^2+xy+y^2=1$
$\displaystyle (x+y)^2-xy=1 \Rightarrow(x+y)^2-1^2=xy$
$\displaystyle (x+y+1)(x+y-1)=xy$
Isomorphism: Whatever follows is wrong.
Case 1: $\displaystyle x+y+1=x$, $\displaystyle x+y-1=y$,
$\displaystyle y=-1,x=1$
Case 2: $\displaystyle x+y+1=y$, $\displaystyle x+y-1=x$,
$\displaystyle x=-1,y=1$

Only possible value of $\displaystyle xy$ is $\displaystyle -1$

First post with latex.
One possible line of attack:

$\displaystyle x^2 + xy + y^2 = 1$

See it as a quadratic in x, then $\displaystyle x = \dfrac{-y \pm \sqrt{4 - 3y^2}}{2}$

Now for a real solution to exist, $\displaystyle 4 \ge 4 - 3y^2 \ge 0$
Thus $\displaystyle y^2 \le \frac{4}{3}$

And then $\displaystyle xy = \dfrac{-y^2 \pm \sqrt{4y^2 - 3y^4}}{2}$.....

3. ## Question

I was wondering if that method was correct...
May I ask what makes this method wrong in this case, I've seen it used in other questions. (Is it that two variables are being related to each other and not to a real number?)

Also, does this mean there is an infinite number if values of $\displaystyle xy$ when both $\displaystyle x$and $\displaystyle y$are real numbers?

4. Originally Posted by I-Think
I was wondering if that method was correct...
May I ask what makes this method wrong in this case, I've seen it used in other questions. (Is it that two variables are being related to each other and not to a real number?)
Here's a counterexample: $\displaystyle x = 0, y = 1$,then $\displaystyle (x+y+1)(x+y-1)=xy$. However neither of your cases are true!!

Also, does this mean there is an infinite number if values of $\displaystyle xy$ when both $\displaystyle x$and $\displaystyle y$are real numbers?
Yes, I think they asked for a range of values...