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ICSE Class X Math 2015 Paper

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Question : 51 of 52
Marks: +1, -0
Given   x3+12x6x2+8=  y3+27y9y2+27.    \; \frac{x^3+12x}{6x^2+8} = \; \frac{y^3+27y}{9y^2+27} . \; \; Using componendo and dividendo find x:yx: y.
Solution:  
  x3+12x6x2+8=  y3+27y9y2+27.    \; \frac{x^3+12x}{6x^2+8} = \; \frac{y^3+27y}{9y^2+27} . \; \;
Applying componendo and dividendo
  x3+12x+6x2+8x3+12x−6x2−8  \; \frac{x^3+12x+6x^2+8}{x^3+12x-6x^2-8} \; =  y2+27y+9y2+27y3+27y−9y2−27= \; \frac{y^2+27y+9y^2+27}{y^3+27y-9y^2-27}
  (x+2)3(x−2)3  =  (y+3)3(y−3)3\; \frac{(x+2)^3}{(x-2)^3} \; = \; \frac{(y+3)^3}{(y-3)^3}
  x+2x−2  =  y+3y−3\; \frac{x+2}{x-2} \; = \; \frac{y+3}{y-3}
Again, applying componendo and dividendo
  x+2+x−2x+2−x+2  =  y+3+y−3y+3−y+3\; \frac{x+2+x-2}{x+2-x+2} \; = \; \frac{y+3+y-3}{y+3-y+3}
  2x4  =  2y6\; \frac{2x}{4} \; = \; \frac{2y}{6}
  xy  =  46=  23\; \frac{x}{y} \; = \; \frac{4}{6} = \; \frac{2}{3}
x:y  =2:3.x: y \; = 2: 3 .
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