Question

If 𝑥 ≠ 𝑦 and \[\frac{𝑥^2−𝑥+1}{𝑦^2−𝑦+1} = \frac{𝑥^2+𝑥+1}{𝑦^2+𝑦+1}\], prove that 𝑥⁢𝑦 = 1. Use properties of proportion.

Answer 1

Given,

\[ \frac{x^2-x+1}{y^2-y+1} = \frac{x^2+x+1}{y^2+y+1} \]

Applying alternendo we get,

\[ \frac{x^2-x+1}{x^2+x+1} = \frac{y^2-y+1}{y^2+y+1} \]

Applying componendo and dividendo we get,

\[ \Rightarrow \frac{(x^2-x+1)+(x^2+x+1)} {(x^2-x+1)-(x^2+x+1)} = \frac{(y^2-y+1)+(y^2+y+1)} {(y^2-y+1)-(y^2+y+1)} \]

\[ \Rightarrow \frac{x^2-x+1+x^2+x+1} {x^2-x+1-x^2-x-1} = \frac{y^2-y+1+y^2+y+1} {y^2-y+1-y^2-y-1} \]

\[ \Rightarrow \frac{2x^2+2}{-2x} = \frac{2y^2+2}{-2y} \]

\[ \Rightarrow \frac{2(x^2+1)}{-2x} = \frac{2(y^2+1)}{-2y} \]

\[ \Rightarrow \frac{x^2+1}{x} = \frac{y^2+1}{y} \]

\[ \Rightarrow xy^2+x=x^2y+y \]

\[ \Rightarrow xy(y-x)-(y-x)=0 \]

\[ \Rightarrow (y-x)(xy-1)=0 \]

\[ \Rightarrow xy=1 \]

Hence, proved that \(xy=1\).