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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.

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Ok, so you've got $$xy+yz+zx=0$$

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

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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 $xyz=-\dfrac{d}{a}=-112$.

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