How many ordered triplet solutions $(x,y,z)$ does the system $x+y+z=4$, $x^2+y^2+z^2=14$ and $x^3+y^3+z^3=34$ have
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$(x+y+z)^2=x^2+y^2+z^2+2(xy+yz+xz)$ – Samar Imam Zaidi Dec 27 '19 at 03:42
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$x^3 + y^3 + z^3 – 3xyz = (x + y + z) (x^2 + y^2 + z^2 – xy – yz – zx)$ – Samar Imam Zaidi Dec 27 '19 at 03:48
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Since there are $3$ variables and $3$ independent equations, we have atmost $1$ solution upto ordering. Since $(3,2,-1)$ satisfy these, this is the solution and the number of ordered triplets are $3!=6$.
Martund
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