By using the Master Theorem which is a  tool used to analyze the asymptotic behavior of divide-and-conquer algorithm; What is the complexity of an algorithm described by the recurrence T(n)=4T(n/3)+n2.
The algorithm described by the recurrence T(n) = 4T(n/3) + n^2 has a time complexity of O(n^3).

To determine this, we can use the Master Theorem, which states that if a recurrence relation has the form T(n) = aT(n/b) + f(n), then the time complexity is given by the following cases:

If f(n) = O(n^logb_a), then T(n) = O(n^logb_a).
If f(n) = Θ(n^logb_a * log n), then T(n) = Θ(n^logb_a * log n).
If f(n) = Ω(n^logb_a) and a > b^c (c > 0), then T(n) = Θ(n^logb_a).

In this case, we have a = 4, b = 3, and f(n) = n^2. We can calculate log3_4, which is approximately 0.79. Since n^2 = Ω(n^log3_4), we can conclude that this algorithm falls under case 3 of the Master Theorem, which means that its time complexity is T(n) = Θ(n^3).