solution of the cubic equation $x^3 = 0$? it is clear that one solution of the above equation is $x = 0$, but since it is a cubic equation so it should have $3$ complex roots, so what will be its other roots? Also how can we determine that a cubic equation has $3$ distinct roots or $2$ same $1$ distinct or all distinct roots, i.e. how can we predict the nature of roots for a cubic equation? Please guide me.
 A: In the case of $x^3 = 0$, there is only one root. If $x \neq 0$, then we can divide both sides by $x$ to get $x^2 = 0/x = 0$, and similarly, $x = 0/x = 0$, a contradiction.
For a general cubic, the most elementary way to determine whether roots are repeated is to differentiate the cubic. For a cubic $p$ and a point $a \in \Bbb{C}$, $p(a)=p'(a)=0$ if and only if $a$ is a repeated root of $p$. Since quadratics are easy enough to solve with the quadratic formula, you can solve $p'(x) = 0$, then plug the resulting solutions into $p$. If at least one is zero, then you have a repeated root. If you don't get zero, then you have no repeated roots!
If $p'(a) = p(a) = 0$ and $p'$ has a repeated root at $a$, then $a$ is a root of $p$ with multiplicity $3$. That is, $p$ is a non-zero multiple of $(x - a)^3$.
For example, consider the cubic polynomial
$$p(x) = x^3 - 3x^2 + 4.$$
We solve the derivative equal to $0$:
$$p'(x) = 3x^2 - 6x = 3x(x - 2) = 0 \implies x = 0 \text{ or } x = 2.$$
We then see that $p(0) = 4 \neq 0$, but $p(2) = 0$. This means, $2$ is a repeated root of $p$. It only has multiplicity $2$, otherwise we would expect $p'$ to have a repeated root at $2$ as well. Thus, we have one repeated root, and one other distinct root.

To answer your question in the comments, you can determine with derivatives, in the case of $3$ distinct roots, whether only one is real or whether all three are real. The method isn't as clean as the discriminant, but it works.
Find the solutions to $p'(x) = 0$, i.e. the stationary points of $p$. If $a$ is a repeated root of $p'(x) = 0$ and $p(a) = 0$, then we already know that we get a real root of multiplicity $3$.
On the other hand, if $p(a) \neq 0$, but $a$ is still a repeated root of $p'(x) = 0$, then $p$ has one real root, and two non-real roots.
Otherwise, we have two distinct roots for $p'$. Test, using the first or second derivative test, the nature of these stationary points. We'll call a stationary point $a$ "bad" if it is either


*

*A local minimum, and $p(a) > 0$, or

*A local maximum, and $p(a) < 0$.


If we have a "bad" stationary point, then we only have one real root. Otherwise, we have three real roots.
A: First, 
$$  (x-0)(x-0)(x-0) = x^3  $$
exhibits the three roots of $x^3$.  They're all $0$.
Much like the discriminant of the quadratic, the discriminant of the cubic indicates the types of the roots.  The polynomial $x^3$ has discriminant $0$ so at least two roots are the same.  Generally, if the coefficients of a cubic are real numbers and the discriminant is not zero, the discriminant is positive if the roots are three distinct real numbers, and negative if there is one real root and two complex conjugate roots.
Using that $x^3$ has discriminant zero, we write 
$$  x^3 = (x-a)(x-a)(x-b)  $$
where $a$ is a definitely repeated root and $b$ may be different or the same as $a$.  We multiply out, obtaining
$$  x^3 = x^3 - (2a+b) x^2 + (a^2 + 2 a b) x - a^2b  \text{.}  $$
You should have been exposed to the fact that two polynomials are the same only if the coefficients are the same, so comparing the coefficient on the right and left, we get the system
$$  \begin{cases}
x^3 :&  1 = 1  \\
x^2 :&  0 = -(2a+b)  \\
x   :&  0 = a^2 + 2ab  \\
x^0 :&  0 = -a^2 b
\end{cases}
$$
There are several ways to solve this system, but the fastest I see is this:
Multiply the $x^2$ equation by $-a$, obtaining
$$  0 = 2a^2 + 2ab  \text{,}  $$
subtract the $x$ equation from this, obtaining
$$  0 = a^2  \text{,}  $$
which forces $a = 0$.  Then either the $x$ or $x^2$ equation gives $b = 0$.  So all three roots are $0$.
A: The claim that a complex polynomial of degree $n$ has $n$ roots is only true when the roots are counted with multiplicity. In your case, the polynomial $x^3$ has the root $0$ of multiplicity $3$. The polynomial $(x-1)^2(x-2)$ for example has the root $1$ of multiplicity $2$ and the root $2$ of multiplicity $1$. It is the sum of multiplicities of all of the roots that is equal to the degree of the polynomial, not the number of distinct roots per se.
Indeed, there is a root $a$ of the polynomial if and only if the factorisation of the polynomial contains a term of the form $(x-a)^k$. The exponent $k$ here is the multiplicity of this root. So to determine the nature of the roots of the polynomial, what we need to do is to look at the factorisation.
A: Solving this type of equation is very simple because you can always rewritw it as follows:
$$(x-0)^3=0$$
From here, you can see that the only solution to the equation is:
$$x=0$$
because being $x$ a general complex numer when $x\neq 0$ you have that $x^3\neq 0$.
