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Mathematics

If x2+y2x2−y2=218\dfrac{x^2 + y^2}{x^2 - y^2} = 2\dfrac{1}{8}, find :

(i) xy\dfrac{x}{y}

(ii) x3+y3x3−y3\dfrac{x^3 + y^3}{x^3 - y^3}

Ratio Proportion

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Answer

Given,

⇒x2+y2x2−y2=218⇒x2+y2x2−y2=178\Rightarrow \dfrac{x^2 + y^2}{x^2 - y^2} = 2\dfrac{1}{8} \\[1em] \Rightarrow \dfrac{x^2 + y^2}{x^2 - y^2} = \dfrac{17}{8}

Applying componendo and dividendo:

⇒x2+y2+x2−y2x2+y2−(x2−y2)=17+817−8⇒2x22y2=259⇒x2y2=259⇒xy=53.\Rightarrow \dfrac{x^2 + y^2 + x^2 - y^2}{x^2 + y^2 - (x^2 - y^2)} = \dfrac{17 + 8}{17 - 8} \\[1em] \Rightarrow \dfrac{2x^2}{2y^2} = \dfrac{25}{9} \\[1em] \Rightarrow \dfrac{x^2}{y^2} = \dfrac{25}{9} \\[1em] \Rightarrow \dfrac{x}{y} = \dfrac{5}{3}.

Hence, xy=53=123\dfrac{x}{y} = \dfrac{5}{3} = 1\dfrac{2}{3}.

(ii) We know that,

⇒xy=53⇒x3y3=12527⇒x3+y3x3−y3=125+27125−27⇒x3+y3x3−y3=15298⇒x3+y3x3−y3=7649⇒x3+y3x3−y3=12749.\phantom{\Rightarrow} \dfrac{x}{y} = \dfrac{5}{3} \\[1em] \Rightarrow \dfrac{x^3}{y^3} = \dfrac{125}{27} \\[1em] \Rightarrow \dfrac{x^3 + y^3}{x^3 - y^3} = \dfrac{125 + 27}{125 - 27} \\[1em] \Rightarrow \dfrac{x^3 + y^3}{x^3 - y^3} = \dfrac{152}{98} \\[1em] \Rightarrow \dfrac{x^3 + y^3}{x^3 - y^3} = \dfrac{76}{49} \\[1em] \Rightarrow \dfrac{x^3 + y^3}{x^3 - y^3} = 1\dfrac{27}{49}.

Hence, x3+y3x3−y3=12749\dfrac{x^3 + y^3}{x^3 - y^3} = 1\dfrac{27}{49}.

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