The function f(x)=x^3+ax^2+bx+c has a relative maximum at (-3, 25) and a point of inflection at x=-1. find a,b, and c?

f(x) = x³ + ax² + bx + c
f'(x) = 3x² + 2ax + b
f"(x) = 6x + 2a

Since there is a point of inflection at x = -1, then f"(-1) = 0
6(-1) + 2a = 0
a = 3
and f'(x) = 3x² + 6x + b

Since there is a point relative maximum at x = -3, then f'(-3) = 0
3(-3)² + 6(-3) + b = 0
b = -9
Then, f(x) = x³ + 3x² - 9x + c

(-3, 25) lies on the graph of f(x), then f(-3) = 25
(-3)³ + 3(-3)² - 9(-3) + c = 25
c = -2

f(x) = x³ + 3x² - 9x - 2