If for a non-zero x, 3$$x^{2}+5x + 3 = 0$$, then the value of $$\ x^{3}+\frac{1}{x^{3}}$$ is:

Given : $$3x^2+5x+3=0$$

Divide by $$x$$, => $$3x+\frac{3}{x}=-5$$

=> $$x+\frac{1}{x}=\frac{-5}{3}$$ -------------(i)

Cubing both sides, we get :

=> $$(x+\frac{1}{x})^3=(\frac{-5}{3})^3$$

=> $$x^3+\frac{1}{x^3}+3(x)(\frac{1}{x})(x+\frac{1}{x})=\frac{-125}{27}$$

=> $$x^3+\frac{1}{x^3}+3(1)(x+\frac{1}{x})=\frac{-125}{27}$$

=> $$x^3+\frac{1}{x^3}+3(\frac{-5}{3})=\frac{-125}{27}$$

=> $$x^3+\frac{1}{x^3}=\frac{-125}{27}+5$$

=> $$x^3+\frac{1}{x^3}=\frac{-125+135}{27}$$

=> $$x^3+\frac{1}{x^3}=\frac{10}{27}$$

=> Ans - (A)