Question

Х+y+z = x^2 +y^2 +z^2 Xyz =1

Answer 1

сложно и длинно получилось :

Answer 2

To find the values of variables $x$, $y$, and $z$ that satisfy the given equations:

1. $x + y + z = x^2 + y^2 + z^2$
2. $xyz = 1$

We can use a combination of algebraic techniques to solve the system of equations.

First, let's simplify equation (1):

$x + y + z = x^2 + y^2 + z^2$

Move all terms to one side of the equation:

$x^2 - x + y^2 - y + z^2 - z = 0$

Now, we need to find the sum of the squares and the sum of the terms individually:

$x^2 - x = x(x-1)$ $y^2 - y = y(y-1)$ $z^2 - z = z(z-1)$

So the equation can be rewritten as:

$x(x-1) + y(y-1) + z(z-1) = 0$

Next, let's examine equation (2):

$xyz = 1$

Now, we can try different combinations of values for $x$, $y$, and $z$ that satisfy both equations.

Let's consider a possible solution where $x = 1$, $y = 1$, and $z = 1$:

$1(1-1) + 1(1-1) + 1(1-1) = 0$

$1 \cdot 0 + 1 \cdot 0 + 1 \cdot 0 = 0$

This solution satisfies equation (1) but not equation (2) since $xyz = 1 \cdot 1 \cdot 1 = 1$, which doesn't match the condition $xyz = 1$.

Let's try another combination where $x = -1$, $y = 1$, and $z = 1$:

$-1(-1-1) + 1(1-1) + 1(1-1) = 0$

$-1 \cdot -2 + 1 \cdot 0 + 1 \cdot 0 = 0$

$2 + 0 + 0 = 0$

This solution also satisfies equation (1) but not equation (2) since $xyz = -1 \cdot 1 \cdot 1 = -1$, which doesn't match the condition $xyz = 1$.

After trying various combinations, we can conclude that there are no real values of $x$, $y$, and $z$ that simultaneously satisfy both equations:

$x + y + z = x^2 + y^2 + z^2$

$xyz = 1$

This means there is no real solution for the system of equations as stated.