An algebra problem by Priyanshu Mishra

$\begin{aligned} \Large \prod_{n=2}^{\infty} \frac{\color{#D61F06}{n^3 -1}}{\color{#3D99F6}{n^3+1}} & \Large = \prod_{n=2}^{\infty} \frac{\color{#D61F06}{n-1}}{\color{#3D99F6}{n+1}} \times \prod_{n=2}^{\infty} \frac{\color{#D61F06}{n^2 + n+1}}{\color{#3D99F6}{n^2 -n-1}} & \large \color{#D61F06}{n^3 - 1 = (n-1)(n^2 +n+1)} \color{#333333}{,} \ \large \color{#3D99F6}{ n^3 + 1 = (n+1)(n^2-n+1)} \\ & \Large = {\left(\frac{1}{{\color{#D61F06} \cancel{3}}} \times \frac{2}{{\color{#3D99F6}\cancel{4}}} \times \frac{\color{#D61F06}{\cancel{3}}}{\color{#D61F06}{\cancel{5}}} \times \frac{\color{#3D99F6}{\cancel{4}}}{\color{#3D99F6}{\cancel{6}}} \times \frac{\color{#D61F06}{\cancel{5}}}{\color{#D61F06}{\cancel{7}}} \times ... \right)} \times {\left(\frac{\color{#D61F06}{\cancel{7}}}{3} \times \frac{\color{#3D99F6}{\cancel{13}}}{{\color{#D61F06}\cancel{7}}} \times \frac{\color{#D61F06}{\cancel{21}}}{\color{#3D99F6}{\cancel{13}}} \times \frac{\color{#3D99F6}{\cancel{31}}}{\color{#D61F06}{\cancel{21}}} \times \frac{\color{#D61F06}{\cancel{43}}}{\color{#D61F06}{\cancel{31}}} \times ... \right)} & \large \color{#3D99F6}{\text{By using Telescoping Series}}\\ & \Large = \large 2 \times \frac{1}{3} \\ & \Large = \large \color{#69047E}{\boxed{\frac{2}{3}}} \\ \end{aligned}$