If $$\frac{x^{3}+3y^{2}x}{y^{3}+3x^{2}y}=\frac{35}{19}$$, what is $$\frac{x}{y}$$=

Given : $$\frac{x^{3}+3y^{2}x}{y^{3}+3x^{2}y}=\frac{35}{19}$$

By componendo and dividendo,

=> $$\frac{x^{3}+3y^{2}x+(y^3+3x^2y)}{x^{3}+3y^{2}x-y^3-3x^2y}=\frac{35+19}{35-19}=\frac{54}{16}$$

=> $$\frac{(x+y)^3}{(x-y)^3}=\frac{27}{8}=(\frac{3}{2})^3$$

=> $$\frac{x+y}{x-y}=\frac{3}{2}$$

By componendo and dividendo again, we get :

=> $$\frac{x+y+x-y}{x+y-x+y}=\frac{3+2}{3-2}$$

=> $$\frac{x}{y}=5$$

=> Ans - (C)