Question

If $\frac{u}{v}=\frac{w}{x}=\frac{y}{z}$ then show that each ratio is equal to ${\left[\frac{u{w}^2+7u{y}^2+3y{u}^2}{u{x}^2+7x^2+3z{v}^2}\right]}^{\frac{1}{3}}.$

Answer 1

$$\begin{gathered} Let: \\ \frac{u}{v}=\frac{w}{x}=\frac{y}{z}=k \\ Thus,wehave: \\ u=vk,w=xkandy=zk \\ Now,c\mathrm{onsider}{\left[\frac{u{w}^2+7w{y}^2+3y{u}^2}{v{x}^2+7x{z}^2+3z{v}^2}\right]}^{\frac{1}{3}}. \end{gathered}$$

<sup>$\mathrm{Substituting}\mathrm{the}\mathrm{values}\mathrm{of}u,v\mathrm{and}y,weget:$

${\left[{\displaystyle \frac{v{x}^2{k}^3+7x{z}^2{k}^3+3zv{k}^3}{v{x}^2+7x{z}^2+3z{v}^2}}\right]}^{\frac{1}{3}}$
$={\left(k^3\right)}^{\frac{1}{3}}$
$=k$</sup>
We know that $\frac{u}{v}=\frac{w}{x}=\frac{y}{z}=k$.
Hence, each ratio is equal to ${\left[\frac{u{w}^2+7w{y}^2+3y{u}^2}{v{x}^2+7x{z}^2+3z{v}^2}\right]}^{\frac{1}{3}}$.