If {a^x} = bc,{b^y} = ca,and {c^z} = ab,then | vedclass

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(D) Given equations are:$a^x = bc$  --- $(1)$$b^y = ca$  --- $(2)$$c^z = ab$  --- $(3)$Taking logarithm on both sides of each equation:$x \log a = \lo

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$x + y + z + 2$

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(D) Given equations are:$a^x = bc$  --- $(1)$$b^y = ca$  --- $(2)$$c^z = ab$  --- $(3)$Taking logarithm on both sides of each equation:$x \log a = \log b + \log c$  --- $(4)$$y \log b = \log c + \log a$  --- $(5)$$z \log c = \log a + \log b$  --- $(6)$Let $A = \log a$,$B = \log b$,and $C = \log c$.The equations become:$xA = B + C$$yB = C + A$$zC = A + B$From these,$x = \frac{B+C}{A}$,$y = \frac{C+A}{B}$,$z = \frac{A+B}{C}$.Multiplying these,$xyz = \frac{(B+C)(C+A)(A+B)}{ABC}$.This does not simplify to a constant unless specific values are assumed. However,checking the standard identity for this problem type where $x+1 = \frac{A+B+C}{A}$,etc.,the expression $xyz$ is usually related to $x+y+z+2$. Specifically,$(x+1)(y+1)(z+1) = (\frac{A+B+C}{A})(\frac{A+B+C}{B})(\frac{A+B+C}{C}) = \frac{(A+B+C)^3}{ABC}$.Given the options provided,the correct algebraic identity derived from these equations is $xyz + x + y + z + 2 = (x+1)(y+1)(z+1) - 1$. Actually,for the system $a^x=bc$,etc.,the identity is $xyz = x+y+z+2$.

If ${a^x} = bc$,${b^y} = ca$,and ${c^z} = ab$,then find the value of $xyz$.

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