<Li> If p and q are real numbers and q / 4 + p / 27 <0, then the numbers (4) and (5) are complex numbers each of which is the conjugate of the other one . Let u be a cube root of (4) and let v be the conjugate of u . Then v is the conjugate of u and this proves that v is equal to (5). So, again, since u × v is real we have u × v = − p / 3, and therefore the roots of the equation (2) are u + u, ξu + ξu, and ξu + ξu . In this case all roots are real, since each one of them is the sum of a complex number with its conjugate . </Li> <Li> If p = 0, then the roots of the equation (2) are the cube roots of − q . This is compatible with Cardano's formula, because one of (4) or (5) is 0 and the other is − q . </Li> <Li> If q = 0, then the roots of (2) are 0 and the square roots of − p . Again, this is compatible with Cardano's formula, because if u is a square root of ⁄, then u is a square root of p / 27, and this square root is equal to (4) or to (5), since we are assuming that q = 0 . If v = − u is the other square root of ⁄ then, by the same reason, v is equal to (4) or to (5) and furthermore if u is equal to (4) then v is equal to (5) and vice versa . Because v = − u we have u + v = 0 . On the other hand, ξu + ξv = (ξ − ξ) × u and so (ξu + ξv) = (ξ − ξ) × u = (− 3) × p / 3 = − p, which means that ξu + ξv is a square root of − p . Finally, ξu + ξv = (ξ − ξ) × u = − (ξu + ξv), and so it must be the other square root of − p . </Li> <Li> If q / 4 + p / 27 = 0 (but p and q are not 0), then (2) has a simple root, which is 3q / p, and a double root, which is − 3q / 2p . Again, this is compatible with Cardano's formula . To see why, note that asserting that q / 4 + p / 27 = 0 is equivalent to asserting that 27q / 4p = − 1 . If u = v = 3q / 2p, then u = v = 27q / 8p = − q / 2 and 3uv = 27q / 4p = − p . Thus, Cardano's formula says that the roots of (2) are u + v = 2u = 3q / p, ξu + ξv = (ξ + ξ) × u = − 3q / 2p, and ξu + ξv = (ξ + ξ) × u = − 3q / 2p . </Li>

How many unique roots will a third degree polynomial function have