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.

Ok, so you've got $$xy+yz+zx=0$$
Now, we know that $$x^3+y^3+z^3-3xyz=(x+y+z)(x^2+y^2+z^2-xy-yz-zx)\\
\implies7-3xyz=7(49-0)\\
\implies xyz=-\dfrac{7\times48}{3}=-112$$
