If {a^{x - 1}} = bc,{b^{y - 1}} = ca,and {c^{ | vedclass

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(A) Given: ${a^{x - 1}} = bc \Rightarrow {a^x} = abc$ Similarly,${b^{y - 1}} = ca \Rightarrow {b^y} = abc$ and ${c^{z - 1}} = ab \Rightarrow {c^z} = a

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(A) Given: ${a^{x - 1}} = bc \Rightarrow {a^x} = abc$ Similarly,${b^{y - 1}} = ca \Rightarrow {b^y} = abc$ and ${c^{z - 1}} = ab \Rightarrow {c^z} = abc$. Let ${a^x} = {b^y} = {c^z} = abc = k$. Then $a = {k^{1/x}}$,$b = {k^{1/y}}$,and $c = {k^{1/z}}$. Taking the product: $abc = {k^{1/x}} \cdot {k^{1/y}} \cdot {k^{1/z}} = {k^{(1/x + 1/y + 1/z)}}$. Since $abc = k$,we have $k^1 = {k^{(1/x + 1/y + 1/z)}}$. Equating the exponents: $1 = \frac{1}{x} + \frac{1}{y} + \frac{1}{z} = \sum \frac{1}{x}$.

If ${a^{x - 1}} = bc$,${b^{y - 1}} = ca$,and ${c^{z - 1}} = ab$,then find the value of $\sum \frac{1}{x}$.

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