# absolute and Cubic System of equations of $3$ variables

(1) Total no. of ordered pairs that satisfy the system of equations.

$x^2-\mid x \mid =yz\;\;$

$y^2-\mid y \mid = zx\;\;$

$z^2-\mid z\mid =xy$

(2) Total no. of ordered pairs that satisfy the system of equations.

$y^3-6x^2+12x-8 = 0$

$z^3-6y^2+12y-8 = 0$

$x^3-6z^2+12z-8 = 0$

My try: for first (1)

Clearly $x=y=z=0$ are the solution of Given equations.

Now If $x\neq 0 ,y\neq 0 ,z\neq 0$, Then

$x^2-\mid x \mid \leq x^2\Leftrightarrow yz\leq x^2$

$y^2-\mid y \mid \leq y^2\Leftrightarrow zx\leq y^2$

$z^2-\mid z \mid \leq x^2\Leftrightarrow xy\leq z^2$

........................................

$x^2y^2z^2\leq x^2y^2z^2$

which is possible only when all are equal or at least one is zero.

I did not understand after that plz explain me

(2)

$y^3 = 6x^2-12x+8\geq 2\Rightarrow y\in \left[\sqrt[3]{2},\infty\right)$

$z^3 = 6y^2-12y+8\geq 2\Rightarrow z\in \left[\sqrt[3]{2},\infty\right)$

$x^3 = 6z^2-12z+8\geq 2\Rightarrow z\in \left[\sqrt[3]{2},\infty\right)$

Now Let $f(x)=x^3$(Strictly increasing function in Given Domain)

and $g(x) = 6x^2-12x+8$(Strictly increasing function in Given Domain)

So our System of equation become

$f(y) = g(x)$

$f(z) = g(y)$

$f(x) = g(z)$

Now How can I solve after that

Thanks

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Re 1, what you mean is the equality must hold in all the inequalities. For example, $x=\pm1, y=0, z=0$ satisfies the equations, but not your solution set. –  Calvin Lin Jun 24 '13 at 4:32
Also, please split up your questions ... –  Calvin Lin Jun 24 '13 at 4:34
OP just posted a question from code chef.com I'm not sure if these are similar live contest problems. –  Calvin Lin Jun 24 '13 at 4:39
Again First Question is from Titu Andrescuu book and other question is from problem and analysis G.N Berman. If this question is from code chef Then sorry from my side. I did not know code chef is any site. –  juantheron Jun 24 '13 at 4:58