If {a^x} = {(x + y + z)^y},{a^y} = {(x + y + | vedclass

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(A) Given: ${a^x} = {(x + y + z)^y}$,${a^y} = {(x + y + z)^z}$,${a^z} = {(x + y + z)^x}$.Multiplying the three equations,we get:${a^x} \cdot {a^y} \cd

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$x = y = z = a/3$

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(A) Given: ${a^x} = {(x + y + z)^y}$,${a^y} = {(x + y + z)^z}$,${a^z} = {(x + y + z)^x}$.Multiplying the three equations,we get:${a^x} \cdot {a^y} \cdot {a^z} = {(x + y + z)^y} \cdot {(x + y + z)^z} \cdot {(x + y + z)^x}$${a^{x + y + z}} = {(x + y + z)^{x + y + z}}$This implies $x + y + z = a$.Substituting $x + y + z = a$ into the original equations:${a^x} = {a^y} \Rightarrow x = y$${a^y} = {a^z} \Rightarrow y = z$Since $x = y = z$ and $x + y + z = a$,we have $3x = a$,which means $x = y = z = a/3$.

If ${a^x} = {(x + y + z)^y}$,${a^y} = {(x + y + z)^z}$,and ${a^z} = {(x + y + z)^x}$,then:

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