Let x, y, z be non-zero real numbers such tha | vedclass

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Question

(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^

Vedclass · Accepted Answer

$154$

Vedclass · Answer

(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,

$a^3+b^3+c^3-3 a b c=(a+b+c)$

$\left.	herefore a^3+b^3+b^2+c^2-a b-b c-c aight)$

$\left.\quad=\left[(a+b+c)^2-3(a b+b c+c a)ight)ight]$

$\Rightarrow\left(\frac{x}{y}ight)^3+\left(\frac{y}{z}ight)^3+\left(\frac{z}{x}ight)^3-3=\left(\frac{x}{y}+\frac{y}{z}+\frac{z}{x}ight)$

${\left[\left(\frac{x}{y}+\frac{y}{z}+\frac{z}{x}ight)^2-3\left(\frac{x}{z}+\frac{y}{x}+\frac{z}{y}ight)ight]}$

$	herefore \quad \frac{x^3}{y^3}+\frac{y^3}{z^3}+\frac{z^3}{x^3}-3=(7)\left[7^2-3 	imes 9ight]$

$=7(49-27)=7 	imes 22=154$

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 $\frac{x^3}{y^3}+\frac{y^3}{z^3}+\frac{z^3}{x^3}-3$ is equal to

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