# 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 $$ax^6+bx^3+c=0\;?$$

• Do you allow complex solutions, or just real? – Ross Millikan Jan 12 '13 at 16:38

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 $y_i$.

• Nice illustration! You really pictured the problem. +1 – mrs Jan 15 '13 at 16:34
• Thank you so much for such detailed explanation, I had no internet connection and couldn't reply for some time. My problem was that I didn't realize that $(x^3)^2=x^6$ haha, yeah I am that stupid, so when I saw your answer, starting with "Let $y=x^3$", I immediately realized what the solution would be. So thank you. – Paul Dirac Jan 16 '13 at 10:17

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...

• Simple and encouraging way. +1 – mrs Jan 12 '13 at 17:22

Hint: Set $t=x^3$ and solve for $t$

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 $$\omega=\frac{1}{2}+i\cdot\frac{\sqrt{3}}{2}$$