If $$\tan \left(\alpha+\beta\right)=\surd{3},\tan \left(\alpha-\beta\right)=1$$ where $$\left(\alpha+\beta\right)$$ and $$\left(\alpha-\beta\right)$$ are acute angles, then what is $$\tan \left(6\alpha\right)$$?

A

-1

B

0

C

1

D

$$\surd{2}-1$$