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.

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 $0$.
