If $$x^{2}$$ + (1/$$x^{2}$$) = 31/9 and x > 0, then what is the value of $$x^{3}$$ + (1/$$x^{3}$$) ?

Given : $$x^2+\frac{1}{x^2}=\frac{31}{9}$$

Adding 2 on both sides,

=> $$x^2+\frac{1}{x^2}+2(x)(\frac{1}{x})=\frac{31}{9}+2$$

=> $$(x+\frac{1}{x})^2=\frac{49}{9}$$

=> $$(x+\frac{1}{x})=\sqrt{\frac{49}{9}}=\frac{7}{3}$$ ------------(i)

Cubing both sides, we get :

=> $$(x+\frac{1}{x})^3=(\frac{7}{3})^3$$

=> $$x^3+\frac{1}{x^3}+3(x)(\frac{1}{x})(x+\frac{1}{x})=\frac{343}{27}$$

=> $$x^3+\frac{1}{x^3}+3(\frac{7}{3})=\frac{343}{27}$$

=> $$x^3+\frac{1}{x^3}=\frac{343}{27}-7$$

=> $$x^3+\frac{1}{x^3}=\frac{343-189}{27}=\frac{154}{27}$$

=> Ans - (B)