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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 $\frac{x^{3}}{y^{3}}+\frac{y^{3}}{z^{3}}+\frac{z^{3}}{x^{3}}-3$ is equal to
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Verified Answer
The correct answer is:
$154$
$a^{3}+b^{3}+c^{3}-3 a b c=[a+b+c]\left[(a+b+c)^{2}-3(a b+b c+c a)\right]$
$=[7]\left[(7)^{2}-3(9)\right]$
$=7(49-27)=7 \times 22=154$
$=[7]\left[(7)^{2}-3(9)\right]$
$=7(49-27)=7 \times 22=154$
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