2 equations 3 variables

Given that $x, y, z$ are real, solve (with proof) the system of equations: $$(x-1)(y-1)(z-1) = xyz - 1$$ $$(x-2)(y-2)(z-2) = xyz - 2$$ part b) $x,y,z$ are allowed to be complex.

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Let $$P(t) = (x-t)(y-t)(z-t)$$ . $x,y,z$ are roots of $P(t) = 0$.

Expand and assume

$$P(t) = c - bt + at^2 - t^3$$

The two equations basically say

$$P(1) = c -1$$

and

$$P(2) = c -2$$

The first gives $a = b$

and second gives $2a - b = 3$ and thus $a=b = 3$

Thus $$P(t) = c - 3t + 3t^2 - t^3 = c -1 - (t^3 -3t^2 + 3t - 1) = (c-1) - (t-1)^3$$

Thus $x,y,z$ are roots of 

$$(t-1)^3 = c-1$$

For $x,y,z$ to be real, we need $c=1$ and $x=y=z=1$.

If $x,y,z$ are allowed to be complex, pick a random $c$ and $x,y,z$ are $(c-1) w + 1$ where $w$ are the three cuberoots of unity.