Range of $xyz\;,$ If $x+y+z=4$ and $x^2+y^2+z^2=6$ 
If $x,y,z\in \mathbb{R}$ and $x+y+z=4$ and $x^2+y^2+z^2=6\;,$ Then range of $xyz$

$\bf{My\; Try::}$Using $$(x+y+z)^2=x^2+y^2+z^2+2(xy+yz+zx)$$
So we get $$16=6+2(xy+yz+zx)\Rightarrow xy+yz+zx = -5$$ and given $x+y+z=4$
Now let $xyz=c\;,$ Now leyt $t=x,y,z$ be the roots of cubic equation, Then
$$\displaystyle (t-x)(t-y)(t-z)=0\Rightarrow t^3-(x+y+z)t^2+(xy+yz+zx)t-xyz = 0$$
So we get $\displaystyle t^3-4t^2-5t-c=0$
Now let $f(t)=t^3-4t^2-5t-c\;,$ Then $f'(t)=3t^2-8t-5$
and $f''(t)=6t-8.$ Now for max. and Min.$f'(t)=0\Rightarrow 3t^2-8t-5=0$
So we get $\displaystyle t=\frac{8\pm \sqrt{64+60}}{2\cdot 3}=$
Now How can I solve it after that, Help required, Thanks

By CS inequality we get
$$|x^3+y^3+z^3|\le\sqrt{x^2+y^2+z^2}\sqrt{x^4+y^4+z^4}\tag{1}$$
Where
\begin{align*}
x^4+y^4+z^4&=(x^2+y^2+z^2)^2-2[(xy+yz+xz)^2-2xy^2z-2xyz^2-2x^2yz]\\
&=36-2[5^2-2xyz(4)]\\
x^4+y^4+z^4&=16xyz-14\tag{2}
\end{align*}
Now, from the identity
$$x^3+y^3+z^3-3xyz=(x+y+z)(x^2+y^2+z^2-xy-yz-xz)$$
follows 
\begin{align*}
x^3+y^3+z^3-3xyz&=(4)(6-5)\\
x^3+y^3+z^3&=3xyz+4\tag{3}
\end{align*}
$(1)$, $(2)$ and $(3)$ imply
\begin{align*}
|3xyz+4|\le\sqrt{6}\sqrt{16xyz-14}
\end{align*}
By setting $t=xyz$ and squaring the last inequality we get
$$9t^2+24t+16\le 96t-84\quad\iff\quad 9t^2-72t+100\le0\quad\iff\quad(3t-12)^2-44\le0$$
So 
$$\boxed{\color{blue}{4-\frac23\sqrt{11}\le xyz\le4+\frac23\sqrt{11}}}$$
