If frac{log x}{a^{2}+a b+b^{2}}=frac{log y}{b | vedclass

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(C) Let each ratio be equal to a constant $k$.Then,$\log x = k(a^2 + ab + b^2)$,$\log y = k(b^2 + bc + c^2)$,and $\log z = k(c^2 + ca + a^2)$.Multiply

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(C) Let each ratio be equal to a constant $k$.Then,$\log x = k(a^2 + ab + b^2)$,$\log y = k(b^2 + bc + c^2)$,and $\log z = k(c^2 + ca + a^2)$.Multiplying the first equation by $(a-b)$,we get:$(a-b) \log x = k(a-b)(a^2 + ab + b^2) = k(a^3 - b^3)$.Thus,$\log x^{a-b} = k(a^3 - b^3)$.Similarly,$(b-c) \log y = k(b^3 - c^3)$ and $(c-a) \log z = k(c^3 - a^3)$.Now,let $E = x^{a-b} \cdot y^{b-c} \cdot z^{c-a}$.Taking the logarithm on both sides:$\log E = \log x^{a-b} + \log y^{b-c} + \log z^{c-a}$$\log E = k(a^3 - b^3) + k(b^3 - c^3) + k(c^3 - a^3)$$\log E = k(a^3 - b^3 + b^3 - c^3 + c^3 - a^3) = k(0) = 0$.Since $\log E = 0$,we have $E = 10^0 = 1$ (assuming common log) or $E = e^0 = 1$ (assuming natural log).Therefore,$x^{a-b} \cdot y^{b-c} \cdot z^{c-a} = 1$.

If $\frac{\log x}{a^{2}+a b+b^{2}}=\frac{\log y}{b^{2}+b c+c^{2}}=\frac{\log z}{c^{2}+c a+a^{2}}$,then $x^{a-b} \cdot y^{b-c} \cdot z^{c-a}=$

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