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(C) Let $a = \frac{x}{y}$,$b = \frac{y}{z}$,and $c = \frac{z}{x}$.Given that $a+b+c = 7$ and $\frac{1}{a}+\frac{1}{b}+\frac{1}{c} = 9$.Note that $\fra

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$154$

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(C) Let $a = \frac{x}{y}$,$b = \frac{y}{z}$,and $c = \frac{z}{x}$.Given that $a+b+c = 7$ and $\frac{1}{a}+\frac{1}{b}+\frac{1}{c} = 9$.Note that $\frac{1}{a}+\frac{1}{b}+\frac{1}{c} = \frac{y}{x}+\frac{z}{y}+\frac{x}{z} = 9$.We know the algebraic identity $a^3+b^3+c^3-3abc = (a+b+c)(a^2+b^2+c^2-ab-bc-ca)$.This can be rewritten as $a^3+b^3+c^3-3abc = (a+b+c)((a+b+c)^2-3(ab+bc+ca))$.Here,$abc = (\frac{x}{y})(\frac{y}{z})(\frac{z}{x}) = 1$.Also,$ab+bc+ca = (\frac{x}{y})(\frac{y}{z}) + (\frac{y}{z})(\frac{z}{x}) + (\frac{z}{x})(\frac{x}{y}) = \frac{x}{z} + \frac{y}{x} + \frac{z}{y} = 9$.Substituting these values into the identity:$a^3+b^3+c^3-3(1) = (7)((7)^2 - 3(9))$.$a^3+b^3+c^3-3 = 7(49-27)$.$a^3+b^3+c^3-3 = 7(22) = 154$.

Let $x, y, z$ be non-zero real numbers such that $\frac{x}{y}+\frac{y}{z}+\frac{z}{x}=7$ and $\frac{y}{x}+\frac{z}{y}+\frac{x}{z}=9$. Then,the value of $\frac{x^3}{y^3}+\frac{y^3}{z^3}+\frac{z^3}{x^3}-3$ is equal to

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