Question : If $\frac{x^{3}+3y^{2}x}{y^{3}+3x^{2}y}=\frac{35}{19}$, what is $\frac{x}{y} =?$
Option 1: $\frac{7}{6}$
Option 2: $\frac{5}{6}$
Option 3: $\frac{5}{1}$
Option 4: $\frac{7}{1}$
Correct Answer: $\frac{5}{1}$
Solution :
$\frac{x^{3}+3y^{2}x}{y^{3}+3x^{2}y}=\frac{35}{19}$
Applying componendo and dividendo method,
$\frac{x^{3}+3y^{2}x - y^{3}-3x^{2}y}{x^{3}+3y^{2}x + y^{3}+3x^{2}y}=\frac{35-19}{35+19} = \frac{16}{54}$
⇒ $\frac{(x-y)^3}{(x+y)^3} = \frac{8}{27}$
⇒ $\frac{x-y}{x+y} = \frac{2}{3}$
⇒ $\frac{x+y}{x-y}=\frac{3}{2}$
Again, applying the componendo and dividendo method to solve the equation.
$\frac{x+y+x-y}{x+y-x+y}=\frac{3+2}{3-2}$
$\Rightarrow\frac{x}{y} = \frac{5}{1}$
Hence, the correct answer is $\frac {5}{1}$.
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