An algebra problem by abhishek alva

We can re-write the problem as $a^x=bc,b^y=ca,c^z=ab.$
Multiply all these equations.We get $a^x.b^y.c^z=a^2.b^2.c^2.$
So we can clearly see that $x=y=z=2.$
Therefore $xyz-x-y-z=8-2-2-2=\boxed 2.$

$x=\log_a bc = \log_a b+\log_a c\\ y=\log_b ca = \dfrac{\log_a ca}{\log_a b} = \dfrac{\log_a c + \log_a a}{\log_a b} = \dfrac{1+\log_a c}{\log_a b}\\ z=\log_c ab = \dfrac{\log_a ab}{\log_a c} = \dfrac{\log_a a + \log_a b}{\log_a c} = \dfrac{1+\log_a b}{\log_a c}$

To simplify things, we let $p=\log_a b$ and $q=\log_a c$ . This gives

$x=p+q\\ y=\dfrac{1+q}{p}\\ z=\dfrac{1+p}{q}$

$xyz-x-y-z\\ =(p+q)\left(\dfrac{1+q}{p}\right)\left(\dfrac{1+p}{q}\right) - (p+q) - \dfrac{1+q}{p} - \dfrac{1+p}{q}\\ =\dfrac{(p+q)(1+p)(1+q)}{pq} - \dfrac{pq(p+q)}{pq} -\dfrac{q(1+q)}{pq} - \dfrac{p(1+p)}{pq}\\ =\dfrac{(p+q)(1+p+q+pq) - p^2q-pq^2 - q - q^2 - p - p^2}{pq}\\ =\dfrac{\color{#E81990}{p}\color{teal}{+q}\color{#D61F06}{+p^2}\color{#20A900}{+pq+pq}\color{#EC7300}{+q^2}\color{#3D99F6}{+p^2q}\color{#69047E}{+pq^2}\color{#3D99F6}{ - p^2q}\color{#69047E}{-pq^2}\color{teal}{-q}\color{#EC7300}{-q^2}\color{#E81990}{-p}\color{#D61F06}{-p^2}}{\color{#20A900}{pq}}\\ =\dfrac{\color{#20A900}{2pq}}{\color{#20A900}{pq}}\\ =\boxed{2}$

Simplify xyz, you will clearly see xyz=x+y+z+2, just use basic base exchange property of log to expand.