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> Given that $$x^2+y^2+z^2=49$$ $$x+y+z=x^3+y^3+z^3=7$$
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> Find $xyz$.
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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.