If $$\left(\frac{x}{3}\right) + \left(\frac{3}{x}\right) = 1$$ then the value of $$x^3$$ is?

$$\left(\frac{x}{3}\right) + \left(\frac{3}{x}\right) = 1$$    Eq.(i)

Apply cube on both of the sides.

$$\left[(\frac{x}{3}\right) + \left(\frac{3}{x}\right)]^3 = 1^3$$

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

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

Put Eq.(i) in the above equation.

$$\frac{x^3}{27}+\frac{27}{x^3}+3\times1=1$$

$$\frac{x^3}{27}+\frac{27}{x^3} = 1-3$$
$$\frac{x^3}{27}+\frac{27}{x^3} = -2$$

Now only the possible value of $$x^3$$ will be -27 for satisfying the above-given equation.

So option (a) is the correct answer.