Consider a regular heptagon  $A_1{A}_2{A}_3{A}_4{A}_5{A}_6{A}_7$  inscribed in a unit circle.
Let  $p_1=\left(A_1{A}_2\right)\left(A_1{A}_3\right)\left(A_1{A}_4\right)\left(A_1{A}_5\right)\left(A_1{A}_6\right)\left(A_1{A}_7\right)$,
$p_2=\left(A_2{A}_3\right)\left(A_2{A}_4\right)\left(A_2{A}_5\right)\left(A_2{A}_6\right)\left(A_2{A}_7\right),$

$p_3=\left(A_3{A}_4\right)\left(A_3{A}_5\right)\left(A_3{A}_6\right)\left(A_3{A}_7\right),$

$p_4=\left(A_4{A}_5\right)\left(A_4{A}_6\right)\left(A_4{A}_7\right),$

$p_5=\left(A_5{A}_6\right)\left(A_5{A}_7\right)$ and $p_6=A_6{A}_7$ where $A_i{A}_j$ is length of the segment joining  $A_i$  and  $A_j$ then value of product  ${\left(p_1{p}_2{p}_3{p}_4{p}_5{p}_6\right)}^{\frac{2}{7}}$ is 

Let vertices  $A_1,A_2,\mathrm{....}{A}_7$ are $7^{th}$  roots of unity. Let $A_1\left(1\right),A_2\left(\alpha \right),A_3\left({\alpha }^2\right),A_4\left({\alpha }^3\right)$

$$\begin{gathered} A_5=\left({\alpha }^4\right)=\overline{{\alpha }^3},A_6=\left({\alpha }^5\right)=\overline{{\alpha }^2},A_7=\left({\alpha }^6\right)=\overline{\alpha } \\ (1-\alpha )(1-{\alpha }^2)(1-{\alpha }^3)(1-{\alpha }^4)(1-{\alpha }^5)(1-{\alpha }^6)=7 \\ \Rightarrow {\left(|1-\alpha ||1-{\alpha }^2||1-{\alpha }^3|\right)}^2=7 \\ {p}_1=\left(A_1{A}_2\right)\left(A_1{A}_3\right)\left(A_1{A}_4\right)\left(A_1{A}_5\right)\left(A_1{A}_6\right)\left(A_1{A}_7\right) \\ ={\left(|1-\alpha ||1-{\alpha }^2||1-{\alpha }^3|\right)}^2={\left(\sqrt{7}\right)}^2 \end{gathered}$$

Similarly,$$\begin{gathered} p_2=|1-\alpha ||1-{\alpha }^2||1-{\alpha }^3||1-{\alpha }^2| \\ \sqrt{7}\left(|1-{\alpha }^3||1-a^2|\right) \\ {p}_3=\sqrt{7}\left(|1-{\alpha }^2|\right) \\ {p}_4=\sqrt{7},p_5=|1-\alpha ||1-{\alpha }^2| and p_6=|1-\alpha | \\ \Rightarrow p_1.p_2.p_3.p_4.p_5.p_6={\left(\sqrt{7}\right)}^7 \end{gathered}$$