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Mathematics

If x ≠ y and x2−x+1y2−y+1=x2+x+1y2+y+1\dfrac{x^2 - x + 1}{y^2 - y + 1} = \dfrac{x^2 + x + 1}{y^2 + y + 1}, prove that xy = 1. Use properties of proportion.

Ratio Proportion

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Answer

Given,

x2−x+1y2−y+1=x2+x+1y2+y+1\dfrac{x^2 - x + 1}{y^2 - y + 1} = \dfrac{x^2 + x + 1}{y^2 + y + 1}

Applying alternendo we get,

x2−x+1x2+x+1=y2−y+1y2+y+1\dfrac{x^2 - x + 1}{x^2 + x + 1} = \dfrac{y^2 - y + 1}{y^2 + y + 1}

Applying componendo and dividendo we get,

⇒(x2−x+1)+(x2+x+1)(x2−x+1)−(x2+x+1)=(y2−y+1)+(y2+y+1)(y2−y+1)−(y2+y+1)⇒x2−x+1+x2+x+1x2−x+1−x2−x−1=y2−y+1+y2+y+1y2−y+1−y2−y−1⇒2x2+2−2x=2y2+2−2y⇒2(x2+1)−2x=2(y2+1)−2y⇒x2+1x=y2+1y\Rightarrow \dfrac{(x^2 − x + 1) + (x^2 + x + 1)}{(x^2 − x + 1) − (x^2 + x + 1)} = \dfrac{(y^2 - y + 1)+ (y^2 + y + 1)}{(y^2 - y + 1) - (y^2 + y + 1)} \\[1em] \Rightarrow \dfrac{x^2 − x + 1 + x^2 + x + 1}{x^2 − x + 1 − x^2 - x - 1} = \dfrac{y^2 - y + 1 + y^2 + y + 1}{y^2 - y + 1 - y^2 - y - 1} \\[1em] \Rightarrow \dfrac{2x^2 + 2}{-2x} = \dfrac{2y^2 + 2}{-2y} \\[1em] \Rightarrow \dfrac{2(x^2 + 1)}{-2x} = \dfrac{2(y^2 + 1)}{-2y} \\[1em] \Rightarrow \dfrac{x^2 + 1}{x} = \dfrac{y^2 + 1}{y} \\[1em]

⇒ xy2 + x = x2y + y

⇒ xy(y - x) - (y - x) = 0

⇒ (y − x)(xy − 1) = 0

⇒ xy = 1

Hence, proved that xy = 1.

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