Question

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

• A. $152$
• B. $153$
• C. $154$
• D. $155$

Answer 1

Answer: (C)

$154$

Let $\frac{x}{y}=a,\frac{y}{z}=b,\frac{z}{x}=c,$ then

$a+b+c=\frac{x}{y}+\frac{y}{z}+\frac{z}{x}=7$

$abc=\frac{x}{y}\times \frac{y}{z}\times \frac{z}{x}=1$

$(ab+bc+ca)=\frac{x}{y}\times \frac{y}{z}+\frac{y}{z}\times \frac{z}{x}+\frac{z}{x}\times \frac{x}{y}=\frac{x}{z}+\frac{y}{x}+\frac{z}{y}=9$

$a^3+b^3+c^3-3abc=(a+b+c)\left[\right(a+b+c{)}^2-3(ab+bc+ca)]$

$⟹\frac{x^3}{y^3}+\frac{y^3}{z^3}+\frac{z^3}{x^3}-3$

$=(\frac{x}{y}+\frac{y}{z}+\frac{z}{x})[{(\frac{x}{y}+\frac{y}{z}+\frac{z}{x})}^2-3(\frac{x}{z}+\frac{y}{x}+\frac{z}{y})]$

$=7\cdot \left[\right(7{)}^2-3\left(9\right)]$

$=7(49-27)$

$=7\times 22=154$