Verified answer

Suppose to the contrary that (q₁q₂…q)²+1 were a perfect cube.

Working mod 4, note that any odd prime p = -1 or 1 (mod 4).

Hence, p^2 = 1 (mod 4).

Therefore, (q₁q₂…q)² + 1 = 1 * 1 * ... * 1 + 1 = 2 (mod 4).

However, the cubes mod 4 are 0³ = 0, 1³ = 1, 2³ = 0, and 3³ = 3 (mod 4).

So, we have a contradiction, and so (q₁q₂…q)²+1 can never be a perfect cube.

I hope this helps!