Question

please helpme in this​

Answer 1

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$\mathfrak{The\:Answer\:is}$

(x^3+12x+6x^2+8)/(x^3+12x-6x^2-8) = (y^3+27y+9y^2+27)/(y^3+27y-9y^2-27)[by using componendo and dividendo)

=>

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).

we also know that [a^3+b^3+3a^2b+3ab^2 = (a+b)^3] and [a^3-b^3-3a^2b+3ab^2 = (a-b)^3]

=>

Now we can rewrite the equation as

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

=>

take cube root on both sides we get

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

=>

Now apply componendo and dividendo rule

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

=>

2x/4 = 2y/6

x/y = 2/3

=>

therefore x:y = 2:3

$\boxed{Hope\:This\:Helps}$

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