If ${a^x} = bc,{b^y} = ca,\,{c^z} = ab,$ then $xyz$=
$0$
$1$
$x + y + z$
$x + y + z + 2$
If ${a^x} = {(x + y + z)^y},{a^y} = {(x + y + z)^z}$, ${a^z} = {(x + y + z)^x},$ then
The number of integers $q , 1 \leq q \leq 2021$, such that $\sqrt{ q }$ is rational, and $\frac{1}{ q }$ has a terminating decimal expansion, is
If ${\left( {{2 \over 3}} \right)^{x + 2}} = {\left( {{3 \over 2}} \right)^{2 - 2x}},$then $x =$
If ${a^{1/x}} = {b^{1/y}} = {c^{1/z}}$ and ${b^2} = ac$ then $x + z = $
For $x \ne 0,{\left( {{{{x^l}} \over {{x^m}}}} \right)^{({l^2} + lm + {m^2})}}$${\left( {{{{x^m}} \over {{x^n}}}} \right)^{({m^2} + nm + {n^2})}}{\left( {{{{x^n}} \over {{x^l}}}} \right)^{({n^2} + nl + {l^2})}}=$