Find $xyz$, given that the value of $x^2+y^2+z^2$, $x+y+z=x^3+y^3+z^3=7$ 
Given that $$x^2+y^2+z^2=49$$  $$x+y+z=x^3+y^3+z^3=7$$
Find $xyz$.

My attempt, 
I've used a old school way to try to solve it, but I guess it doesn't work.
I expanded $(x+y+z)^3=x^3+y^3+z^3+3(x^2y+xy^2+x^2z+xz^2+yz^2)+6xyz$ 
Since I know substitute the given information into the equation and it becomes $112=x^2y+xy^2+xz^2+yz^2+2xyz$
In another hand, I also expanded $(x+y+z)^2=x^2+y^2+z^2+2(xy+xz+yz)$
$7^2=49+2(xy+xz+yz)$,
So from here, I know that $xy+xz+yz=0$. 
It seems that I stuck here and don't know how to proceed anymore. 
How to continue from my steps? And is there another trick to solve this question? Thanks a lot.

If we identify $x,y,z$ as the roots of a cubic polynomial $av^3+bv^2+cv+d$ with $a\neq0$ then there are recursions in terms of $a,b,c,d$ for the power sums 
$$P_i:=x^i+y^i+z^i.$$
These are the newton identities. As seen in this link
\begin{align} 
P_0=&+3\\
P_1=&-\dfrac{b}{a}\\
P_2=&-\dfrac{b}{a}P_1-\dfrac{c}{a}2\\
P_3=&-\dfrac{b}{a}P_2-\dfrac{c}{a}P_1-\dfrac{d}{a}P_0\\
\end{align}
or
\begin{align} 
P_1a&+b&=0\\
P_1b&+P_2a+2c&=0\\
P_0d&+P_1c+P_2b+P_3a&=0
\end{align}
We see that these equations are linear in $a,b,c,d$. Inserting $P_0=3,P_1=7,P_2=49,P_3=7$ and rearranging:
\begin{align} \tag1
7a&&+b&&&&&=0\\\tag2
49a&&+7b&&+2c&&&=0\\\tag3
7a&&+49b&&+7c&&+3d&=0
\end{align}
Multiply $(1)$ by $7$ and subtract the result by $(2)$. The result is $-2c=0$ so $c=0$. Multiply $(2)$ by $7$ and subtract the result by $(3)$. The result is $336a+7c-3d=0$. Substituting $c=0$ and dividing by $3$ yields $112a-d=0$. Finally according to the link $xyz=-\dfrac{d}{a}=-112$.
