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.

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 $y_i$.
