If $$\sin^6 \theta + \cos^6 \theta = \frac{1}{3}, 0^\circ < \theta < 90^\circ$$, then what is the value of $$\sin \theta \cos \theta$$?

$$\sin^6\theta+\cos^6\theta=\frac{1}{3}$$

$$\left(\sin^2\theta\right)^3+\left(\cos^2\theta\right)^3=\frac{1}{3}$$

$$\left(\sin^2\theta+\cos^2\theta\right)\left(\sin^4\theta-\sin^2\theta\ \cos^2\theta+\cos^4\theta\ \right)=\frac{1}{3}$$

$$\left(1\right)\left(\sin^4\theta+\cos^4\theta+2\sin^2\theta\ \cos^2\theta-3\sin^2\theta\ \cos^2\theta\ \right)=\frac{1}{3}$$

$$\left(\sin^2\theta+\cos^2\theta\right)^2-3\sin^2\theta\ \cos^2\theta\ =\frac{1}{3}$$

$$1-3\sin^2\theta\ \cos^2\theta\ =\frac{1}{3}$$

$$3\sin^2\theta\ \cos^2\theta\ =\frac{2}{3}$$

$$\sin^2\theta\ \cos^2\theta\ =\frac{2}{9}$$

$$\sin\theta\ \cos\theta=\frac{\sqrt{2}}{3}$$

Hence, the correct answer is Option A