Number of solutions for sixth order equation I have a problem, from Gelfand's "Algebra" textbook, that I've been unable to solve, here it is:
Problem 268. 
What is the possible number of solutions of the equation $$ax^6+bx^3+c=0\;?$$
Thanks in advance.
 A: Hint: $\quad$Let $y = x^3$:
$$ax^6 + bx^3 + c = 0 \quad \iff \quad ay^2 + by + c = 0\tag{1}$$
Solve for $y$ ... there will be either two real solutions, one real solution, or no real solutions when solving for $y$ (why?, when?). (Examine the discriminant.) 


*

*When is $\Delta = \sqrt{b^2 - 4ac}\; < \;0\,$? And what does this mean in terms of the existence (or non-existence), of real solutions in y? 

*When $\Delta = 0$, there is exactly one real-valued solution $y$. 

*When $\Delta > 0$, there are two unique real-valued solutions $y_1, y_2$.
In each case, then, for each (possible) solution $y_i$ of the right-hand equation in $(1)$, what are the number of solutions in $x$ to $y_i = x^3$ for each solution $y_i$?  (Note that the degree $3$ is odd in $y = x^3$, so we don't have to worry whether solutions ($y$'s) are positive or negative. If $y_i$ is a solution, then there will exist $x$ such that $y = x^3$.) Simply check cases for each possible root $y_i$.
A: Put $\,t:=x^3\,$ , so your equation becomes
$$(*) at^2+bt+c=0\Longrightarrow \Delta= b^2-4ac$$
Now, if $\,\Delta=0\,\,$ then $\,(*)\,$ has one unique solution. $\,x^3=t={-b/2a}\,$ , and if $\,\Delta >0\,$ then there're two solutions for $\,t=x^3\,$.
Since $\,3\,$ is an odd natural we don't care whether the solutions above are positive or negative, there always are solutions as long as $\,\Delta\geq0\,$, so now you have to take care of the different cases...
A: Hint: Set $t=x^3$ and solve for $t$
A: The aplication of quadratic formula in
$$
a\cdot (x^3)^2+b\cdot(x^3)+c=0
$$
give us that the possible roots enjoy
$$
x^3 =\left[\frac{-b+\sqrt{b^2+4ac}}{2a}\right]
\quad
\mbox{ or }
\quad
x^{3} =\left[\frac{-b-\sqrt{b^2+4ac}}{2a}\right]
$$
I believe that the greatest difficulty is to extract all the cubic roots of these expressions. For every cubic roots to use  de Moivre's formula.
Extracting the cubics roots of  $\left[\frac{-b+\sqrt{b^2+4ac}}{2a}+ i\cdot 0\right]$ and $\left[\frac{-b-\sqrt{b^2+4ac}}{2a} + i\cdot 0\right]$ by de Moivre's formula we have this equation have six  roots 
$$
x_{+\,i} =\sqrt[3\;]{\frac{-b+\sqrt{b^2+4ac}}{2a}}\cdot \omega^i
\quad
\mbox{ and  }
\quad
x_{-\,i} =\sqrt[3\;]{\frac{-b-\sqrt{b^2+4ac}}{2a}}\cdot \omega^i
$$
where $i=0,1,2$ and $\omega$ is a complex cubic root of unit, for exemple
$$
\omega=\frac{1}{2}+i\cdot\frac{\sqrt{3}}{2}
$$
