If frac{log x}{b-c}=frac{log y}{c-a}=frac{log | vedclass

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(A) Given,$\frac{\log x}{b-c}=\frac{\log y}{c-a}=\frac{\log z}{a-b}=k$ (say). From this,we have $\log x = k(b-c)$,$\log y = k(c-a)$,and $\log z = k(a-

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(A) Given,$\frac{\log x}{b-c}=\frac{\log y}{c-a}=\frac{\log z}{a-b}=k$ (say). From this,we have $\log x = k(b-c)$,$\log y = k(c-a)$,and $\log z = k(a-b)$. This implies $x = e^{k(b-c)}$,$y = e^{k(c-a)}$,and $z = e^{k(a-b)}$. Now,consider the expression $E = x^{b+c} \cdot y^{c+a} \cdot z^{a+b}$. Substituting the values,$E = (e^{k(b-c)})^{b+c} \cdot (e^{k(c-a)})^{c+a} \cdot (e^{k(a-b)})^{a+b}$. Using the property $(e^m)^n = e^{mn}$,we get $E = e^{k(b^2-c^2)} \cdot e^{k(c^2-a^2)} \cdot e^{k(a^2-b^2)}$. Combining the exponents,$E = e^{k(b^2-c^2+c^2-a^2+a^2-b^2)}$. Since the sum in the exponent is $0$,$E = e^{k \cdot 0} = e^0 = 1$.

If $\frac{\log x}{b-c}=\frac{\log y}{c-a}=\frac{\log z}{a-b}$,then the value of $x^{b+c} \cdot y^{c+a} \cdot z^{a+b}$ is

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