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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
$152$
$153$
$154$
$155$
Solution
(c)
Given,
$\Rightarrow \frac{x}{y}+\frac{y}{z}+\frac{z}{x}=7 \Rightarrow \frac{y}{x}+\frac{z}{y}+\frac{x}{z}=9$
We know that,
$a^3+b^3+c^3-3 a b c=(a+b+c)$
$\left.\therefore a^3+b^3+b^2+c^2-a b-b c-c a\right)$
$\left.\quad=\left[(a+b+c)^2-3(a b+b c+c a)\right)\right]$
$\Rightarrow\left(\frac{x}{y}\right)^3+\left(\frac{y}{z}\right)^3+\left(\frac{z}{x}\right)^3-3=\left(\frac{x}{y}+\frac{y}{z}+\frac{z}{x}\right)$
${\left[\left(\frac{x}{y}+\frac{y}{z}+\frac{z}{x}\right)^2-3\left(\frac{x}{z}+\frac{y}{x}+\frac{z}{y}\right)\right]}$
$\therefore \quad \frac{x^3}{y^3}+\frac{y^3}{z^3}+\frac{z^3}{x^3}-3=(7)\left[7^2-3 \times 9\right]$
$=7(49-27)=7 \times 22=154$
