To solve the equation (x+4)(x³-27) = 0, we can set each factor equal to zero and solve for x.

Setting x+4 = 0, we have:
x = -4

Setting x³-27 = 0, we can factor the difference of cubes:
(x-3)(x²+3x+9) = 0

Setting each factor equal to zero, we have:
x-3 = 0 or x²+3x+9 = 0

Solving x-3 = 0, we get:
x = 3

To solve x²+3x+9 = 0, we can use the quadratic formula:
x = (-b ± √(b²-4ac)) / (2a)

In this case, a = 1, b = 3, and c = 9. Plugging these values into the quadratic formula, we get:
x = (-3 ± √(3²-4(1)(9))) / (2(1))
x = (-3 ± √(9-36)) / 2
x = (-3 ± √(-27)) / 2

Since the square root of a negative number is not a real number, there are no real solutions for x in this case.

Therefore, the solutions to the equation (x+4)(x³-27) = 0 are x = -4 and x = 3.