<s>[INST] If $(x+y-7)[z(x+y)+24]=(y+z-7)[x(y+z)+24]=(z+x-7)[y(z+x)+24]$, find $x^2+y^2+z^2$ 
Let x, y, z be pairwise distinct real numbers, if $$(x+y-7)[z(x+y)+24]=(y+z-7)[x(y+z)+24] $$ $$=(z+x-7)[y(z+x)+24]$$, find $x^2+y^2+z^2$

I've tried many ways but couldn't find a working way to solve it. I tried letting $(x+y-7)[z(x+y)+24] = k$ and $t = x + y + z$ but none of these gives useful transformation as far as I can see. Could somebody shed some lights on this? Thanks in advance.
 [/INST] Let $S=x+y+z$. Given is
$$P(x)=(S-x-7)\{x(S-x)+24\}$$
$$=x^3-x^2(2S-7)+x(S^2-7S-24)+24S-168$$
has same value as $P(y), P(z)$.
This means $x,y,z$ are roots of the polynomial, $P(x)+\text{constant}$.
By Vieta's, sum of roots is $S=2S-7 \Rightarrow S=7$. Also
$$xy+yz+zx=-(7^2-7\cdot 7 -24)=24$$
Hence $x^2+y^2+z^2=7^2-2\cdot 24=1$
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