Solutions to $x^3+y^3+z^3 = x^2 + y^2 + z^2=x+y+z=0$

Question

I need to prove that $xyz=1$. dealing trew this problem I get that $x+y= \frac{2}{3}$ and that results that $z = -\frac{2}{3}$. after all i got $xyz= - \frac{10}{27}$

When I was before dealing with this, I over and over get that $xyz=0$.

Answer 1

hint:

Assume $xyz=1$ then $1|xyz$ so $xyz=1a$

Answer 2

You can't prove this for rational numbers x,y,z . There is at least one counter example: $x = y = z = 0$ is a solution which contradicts $xyz = 1$.

Answer 3

Note that for any reasonable (including complex) $x, y, z$ we have $$xy+yz+zx=\cfrac{(x+y+z)^2-x^2-y^2-z^2}{2}=0$$

So we have that $x,y,z$ are roots of the a cubic with unknown constant term $a$, and since the terms in $x$ and $x^2$ have zero coefficient we have

$$x^3+a=0; y^3+a=0; z^3+a=0$$

Adding these together we get $3a=0$, and since $xyz=-a$ we have $xyz=0$, and separately, we conclude $x=y=z=0$.

Answer 4

Use Newton identities to generate a cubic equation that will have $x,y,z$ as roots.

So we'll use the following notation:

$$f(p) = a_3p^3 + a_2p^2 + a_1p + a_0$$

$$s_1 = x + y + z = 0$$ $$s_2 = x^2 + y^2 + z^2 = 0$$ $$s_3 = x^3 + y^3 + z^3 = 0$$

So from Newton Identities we have:

$$a_3s_1 + a_2 = 0$$

For the sake of simplicty we'll use $a_3 = 1$. So we have:

$$a_2 = - s_1 = 0$$

Now we continue, using the Newton Identities:

$$a_3s_2 + a_2s_1 + 2a_1 = 0$$ $$0 + 0 + 2a_1 = 0$$ $$a_1 = 0$$

Now for the last time:

$$a_3s_3 + a_2s_2 + a_1s_1 + 3a_0 = 0$$ $$0 + 0 + 0 + 3a_0 = 0$$ $$a_0 = 0$$

So $f(p) = p^3$.

Now find all roots of it to obtain $x,y,z$.

But we know that $f(p)$ has triple root at $p=0$, so we obtain $x=y=z=0$, and in fact $xyz = 0$

Answer 5

Denote by $e_r$ the elementary symmetric functions and by $p_r$ the power sums of the three complex numbers $x$, $y$, $z$. By Newton's identities it follows from the given assumptions that $$\eqalign{e_1&=p_1=0\ ,\cr e_2&={1\over2}(e_1p_1-p_2)=0\ ,\cr e_3&={1\over3}(e_2p_1-e_1p_2+p_3)=0\ .\cr}$$ Vieta's theorem now says that the three numbers $x$, $y$, $z$ are the three solutions of the equation $Z^3=0$, i.e., that $x=y=z=0$.

Answer 6

$x^2+y^2+z^2=0$ has only the trivial zero solution $(0,0,0)$ as the left hand side is always positive otherwise.