Question

x^3+12x/6x^2+8=y^3+27y/y^2+27 componendo and dividendo find x:y

Answer 1

Hey brother!
I think it's incomplete question
Question is
X^3+12x / 6x^2+8 = Y^3+27y / 9y^2+27
Above question 9 is missing

But above ans is

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Use the componendo and dividendo

(x^3+12x+6x^2+8)/(x^3+12x-6x^2-8) = (y^3+27y+9y^2+27)/(y^3+27y-9y^2-27)

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The numerators are in the form (a^3+b^3+3a^2b+3ab^2) and denominators are in the form (a^3-b^3-3a^2b+3ab^2).


[a^3+b^3+3a^2b+3ab^2 = (a+b)^3] and

[a^3-b^3-3a^2b+3ab^2 = (a-b)^3]

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(x+2)^3/(x-2)^3 = (y+3)^3/(y-3)^3

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take cube root on both sides we get
(x+2)/(x-2) = (y+3)/(y-3)

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Again apply componendo and dividendo rule!


(x+2+x-2)/(x+2-x+2) = (y+3+y-3)/(y+3-y+3)
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=
2x/4 = 2y/6

X/2= y/3
x/y = 2/3

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therefore x:y = 2:3