MCQ
If ${a^x} = bc,{b^y} = ca,\,{c^z} = ab,$ then $xyz$=
- A$0$
- B$1$
- C$x + y + z$
- ✓$x + y + z + 2$
$ \Rightarrow $${a^{x - 2}}{b^{y - 2}}{c^{z - 2}} = 1 = {a^0}{b^0}{c^0}$
$\therefore x = y = z = 2$
$\therefore xyz = {2^3} = 8 = x + y + z + 2$.
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