jmc

Let $S$ be the value of the given expression. Using sum and difference of cubes to factor, we get $$\begin{array}{rl}S& =\frac{(2-1)(2^2+2+1)}{(2+1)(2^2-2+1)}\cdot \frac{(3-1)(3^2+3+1)}{(3+1)(3^2-3+1)}\cdot \frac{(4-1)(4^2+4+1)}{(4+1)(4^2-4+1)}\cdot \frac{(5-1)(5^2+5+1)}{(5+1)(5^2-5+1)}\cdot \frac{(6-1)(6^2+6+1)}{(6+1)(6^2-6+1)}\\ & =\frac{1}{3}\cdot \frac{2}{4}\cdot \frac{3}{5}\cdot \frac{4}{6}\cdot \frac{5}{7}\cdot \frac{2^2+2+1}{2^2-2+1}\cdot \frac{3^2+3+1}{3^2-3+1}\cdot \frac{4^2+4+1}{4^2-4+1}\cdot \frac{5^2+5+1}{5^2-5+1}\cdot \frac{6^2+6+1}{6^2-6+1}.\end{array}$$The first product telescopes to $\frac{1\cdot 2}{6\cdot 7}=\frac{1}{21}$. The second product also telescopes due to the identity $$x^2+x+1=(x+1{)}^2-(x+1)+1.$$That is, the terms $2^2+2+1$ and $3^2-3+1$ cancel, as do the terms $3^2+3+1$ and $4^2-4+1$, and so on, leaving just $\frac{6^2+6+1}{2^2-2+1}=\frac{43}{3}$. Thus, $$S=\frac{1}{21}\cdot \frac{43}{3}=\boxed{\frac{43}{63}}.$$