Find the value of $xyz$ 
$x$, $y$, $z$ are real numbers such that
$①$  $ x+y+z=\sqrt{3}$
$②$ $xy+yz+zx=1$
In this case, find the value of $xyz$.

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Here's my attempt.
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$3=(x+y+z)^2 = x^2 + y^2 + z^2 +  2(xy+yz+zx) = x^2 + y^2 + z^2  + 2$
$\therefore   \ \  x^2 + y^2 + z^2 = 1 $
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$   x^3  +y^3 +  z^3    -   3xyz=(x+y+z)(x^2 +  y^2  +  z^2    -  xy-yz-zx)  = 0$
$\therefore   \ \  x^3 + y^3 + z^3 = 3xyz $
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What can I do... after these?
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I guessed the answer like this.
If $ x=y=z = \frac{\sqrt{3}}{3}$ , then these satisfy $①, ②$.
So, $\  $ $xyz=\frac{\sqrt{3}}{9}$....
Thanks in advance.

If $x,y,z$ are the roots of the equation $x^3 + ax^2 + bx + c = 0$ then $a = -\sqrt 3$ and $b = 1$. 
So, we want $x^3 - \sqrt 3 x^2 + x + c = 0 \equiv C$ to have all real roots. 
This is nicely described by the cubic determinant $$
D_C = a^2b^2 - 4b^3 - 4a^3c - 27c^2 + 18abc
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being greater than or equal to $0$. 
In our case, $D_C  = 3 - 4 + 12\sqrt 3c - 27c^2 - 18\sqrt 3 c = -27c^2 - 6\sqrt 3 c- 1$.
It turns out that $D_C = -27\left(c + \frac 1{3\sqrt 3}\right)^2$, giving $c = \frac {-1}{3 \sqrt 3}$. 
Hence the product of the roots is $-c = \frac{1}{3 \sqrt 3}$.
