Can a cubic equation have three complex roots? I have read somewhere that:

If $x+ iy$ is a root of a cubic equation $ax^3+bx^2+cx+d=0$, then $x-iy$ is also a root of this equation.

My question is:
Can $k-il$ be the other root of this equation if $l\neq 0$?
If yes, then it would seem that $k+ il$ should also be a root, but in that case the cubic equation would have four roots which is impossible.
Where is the mistake in my logic?
($i$ denotes $\sqrt {-1}$, and is sometimes pronounced as $iota$.) 
 A: Your argument is an almost correct proof of the fact that a cubic equation cannot have three complex (non-real) roots. The only thing you fail to account for is multiple roots: you do not treat the case in which the third root is $x+iy$ or $x-iy$; that is, it coincides with one of the other two roots.
This case is impossible for a cubic equation with real $a,b,c,d$, but the theorem you quote is not sufficient to prove this. You need a different argument in this case. The simplest result you can use to exclude it is the following (a special case of Viète's formulas):

if $z_1,z_2,z_3$ (counted with multiplicity) are the roots of a cubic equation $ax^3+bx+cx+d=0$, then $z_1+z_2+z_3 = -\frac{b}{a}$.

A: I’d like to relate something that might be useful to your current level of study, in addition to the other answers.
Start with an observation that might be made to sound pithy but turns out to be quite profound:  given that i and −i are both square roots of −1, how do you know which is which? After all, they share a property that is their defining characteristic!
The answer is that you can’t. Not only is the labeling arbitrary, but you can switch them without changing anything.
So, in general, any time you start with pure real numbers (e.g. the coefficients and constants) and perform normal famialiar arithmetic operations, then any time you wind up with complex numbers, you must have a symmetry such that you can switch all the (i)s and (−i)s without changing anything.
A special case of this is what you’re remembering: that any complex roots of a polynomial must come in pairs (a±bi).
A: If the coefficients $a$, $b$, $c$ and $d$ are real, then the equation $ax^3+bx^2+cx+d=0$, assuming $a\ne0$, has at least a real root: indeed, writing
$$
f(x)=x^3+\frac{b}{a}x^2+\frac{c}{a}x+\frac{d}{a}
$$
we have
$$
\lim_{x\to-\infty}f(x)=-\infty
\qquad
\lim_{x\to\infty}f(x)=\infty
$$
so the intermediate value theorem provides one real value $r$ such that $f(r)=0$.
Since the equation has at most three distinct roots, it follows that it cannot have three distinct complex nonreal roots.
If $a$, $b$, $c$ and $d$ are not supposed to be real, but just complex, then it is surely possible: consider
$$
(x-i)(x-2i)(x-3i)=0
$$
A: Let $z^*$ denote the complex conjugate of z. We have $(z_1z_2)^*=z_1^*z_2^*$ and $(z_1+z_2)^*=z_1^*+z_2^*.$ And $z^*=z$ for real $z.$ Therefore, for real $a,b,c,d$ we have $$0=az^3+bz^2+cz+d\iff 0=0^* =(az^3+bz^2+cz+d)^*=$$ $$=a(z^*)^3+b(z^*)^2+c(z^*)+d.$$ So if   $z_,z_2$ are non-real zeroes of $az^3+bz^2+cz+d ,$ with real $a,b,c,d$   then so are $z_1^*$ and $z_2^*.$ If $z_2\ne \{z_1,z_1^*\},$ this give at least 4 zeroes for the cubic.
