Find the value of $abc-a^2-b^2-c^2$ if $x,y,z$ are given in terms or $a,b,c$ 
if $$x=\frac{a}{b}+\frac{b}{a}$$
$$y=\frac{b}{c}+\frac{c}{b}$$
$$z=\frac{c}{a}+\frac{a}{c}$$
  Then find
  $$abc-a^2-b^2-c^2$$

As there are 6 variables in terms of 3 , we can fix any three without violating the conditions.
Say if $a=b=c=2$ then clearly the answer is$-4$ but how to find it in pure algebric way.
 A: It seems you want to find $xyz-x^2-y^2-z^2.$
If so:
$$xyz-\sum_{cyc}x^2=\prod_{cyc}\left(\frac{a}{b}+\frac{b}{a}\right)-\sum_{cyc}\left(\frac{a}{b}+\frac{b}{a}\right)^2=$$
$$=\frac{\prod\limits_{cyc}(a^2+b^2)}{a^2b^2c^2}-\sum_{cyc}\left(\frac{a^2}{b^2}+\frac{b^2}{a^2}\right)-6=$$
$$=\frac{\sum\limits_{cyc}\left(a^4b^2+b^4a^2+\frac{2}{3}a^2b^2c^2\right)}{a^2b^2c^2}-\sum_{cyc}\left(\frac{a^2}{b^2}+\frac{b^2}{a^2}\right)-6=2-6=-4.$$
In the full writing it's the following:
$$xyz-x^2-y^2-z^2=$$
$$=\frac{(a^2+b^2)(a^2+c^2)(b^2+c^2)}{a^2b^2c^2}-\left(\frac{a^2}{b^2}+\frac{b^2}{a^2}+\frac{a^2}{c^2}+\frac{c^2}{a^2}+\frac{b^2}{c^2}+\frac{c^2}{b^2}+6\right)=$$
$$=\frac{a^4b^2+a^4c^2+b^4a^2+b^4c^2+c^4a^2+c^4b^2+2a^2b^2c^2}{a^2b^2c^2}-$$
$$-\left(\frac{a^2}{b^2}+\frac{b^2}{a^2}+\frac{a^2}{c^2}+\frac{c^2}{a^2}+\frac{b^2}{c^2}+\frac{c^2}{b^2}+6\right)=$$
$$=\left(\frac{a^2}{b^2}+\frac{b^2}{a^2}+\frac{a^2}{c^2}+\frac{c^2}{a^2}+\frac{b^2}{c^2}+\frac{c^2}{b^2}+2\right)-$$
$$-\left(\frac{a^2}{b^2}+\frac{b^2}{a^2}+\frac{a^2}{c^2}+\frac{c^2}{a^2}+\frac{b^2}{c^2}+\frac{c^2}{b^2}+6\right)=2-6=-4.$$
A: As noted in the comments, you can't find $abc-a^2-b^2-c^2$ given $x,y,z$ as stated, since the latter equations imply that $x,y,z$ are algebraically dependent. Indeed, there exists a constant $C$ such that
$$xyz-x^2-y^2-z^2=C\text{.}$$
