Can someone explain this Big-O / Θ confusion? [DSA]

I’m studying DSA and got confused with a basic asymptotic analysis question.

If an algorithm has best case Θ(n) and worst case Θ(n²), why is O(n³) also considered true?

I understand now that Big-O is an upper bound, so if it doesn’t go beyond n², it also doesn’t go beyond n³. But I initially thought O(n³) meant the algorithm actually takes n³ time.

Is the correct way to think about it: Θ = tight bound, O = ceiling/upper bound, Ω = floor/lower bound?

Just trying to make sure I’ve understood this correctly.