If x - (1/x) = 3, then what is the value of $$(2x^{4} + 3x^{3} + 13x^{2} - 3x + 2 / (3x^{4} + 3)?$$

Given : $$x-\frac{1}{x}=3$$ ----------(i)

=> $$\frac{x^2-1}{x}=3$$

=> $$x^2-1=3x$$ ------------(ii)

Squaring equation (i), we get :

=> $$(x-\frac{1}{x})^2=(3)^2$$

=> $$x^2+\frac{1}{x^2}-2(x)(\frac{1}{x})=9$$

=> $$x^2+\frac{1}{x^2}=11$$

=> $$\frac{x^4+1}{x^2}=11$$

=> $$x^4+1=11x^2$$ --------------(iii)

To find : $$\frac{2x^{4} + 3x^{3} + 13x^{2} - 3x + 2}{(3x^{4} + 3)}$$

= $$\frac{2(x^4+1)+3x(x^2-1)+(13x^2)}{3(x^4+1)}$$

Using equations (ii) and (iii),

= $$\frac{2(11x^2)+3x(3x)+13x^2}{3(11x^2)}$$

= $$\frac{22x^2+9x^2+13x^2}{33x^2}$$

= $$\frac{44x^2}{33x^2}=\frac{4}{3}$$

=> Ans - (C)