Solving the general cubic by completing the cube 
Completing the Square:

In a quadratic of the form $x^2+ax+b=0$, we can complete the square by moving $b$ to the right hand side and finding a constant $m$ such that the left hand side becomes a perfect square trinomial. Namely, we have$$m=\left(\frac a2\right)^2\rightarrow \left(x+\frac a2\right)^2=\frac {a^2}4-b\tag{1}$$
Doing so with the general case of $ax^2+bx+c=0$ gives us the quadratic formula$$x=\frac {-b\pm\left(b^2-4ac\right)^{1/2}}{2a}\tag{2}$$


Question:

Just as an "extension" from completing the square, but is it possible to complete the cube? (Get the cubic into the form ($(x+m)^3=\text{something}$).

My Attempt: I expanded out $(x+m)^3$ and got$$(x+m)^3=x^3+3x^2m+3xm^2+m^3\tag{3}$$
After some thought, I realized that the way that I've been doing to complete the cube for $x^3+\frac bax^2+\frac cax+\frac da=0$ was to add $-\frac {2b}{3a}x^2$, $\left(-\frac ca\pm\sqrt{\frac c{3a}}\right)x$, etc. to both sides. But quickly, I realized that doing so would bring the $x$ terms to the right hand side!
So now, I'm wondering if it's possible to even complete the cube and if so, what would the procedures be? (Hints are welcome to!)
Just a Thought: Do you think we'll need the quadratic formula for this?

Actually this can't be solved by making it a cube.it can't be made because then  b and c cannot be arbitrary.there must exist a relation  like $ b^2/3.a^2 = c/a $.   ; if you differentiate, you will get that the cubic equation having only 1 root has exactly that condition.
This happens because when you want to express it as a cubic, you want to  the cube root right!! Then at that point in real plane there is exactly one solution and the other two in complex plane. 
So , here we want to transform in terms of other  variables but in the actual solution though there is some quadratic things popping out .
Neways we can approximate it by writing the solution as nested radicals(quite similarly to the Methods of S.Ramanujan).
If you have anyproblem to understand this ,please feel free to tell me.I will explain it further.
