TITLE: Counting the Number of Real Roots of $y^{3}-3y+1$
QUESTION [13 upvotes]: Here's my question:

How many real roots does the cubic equation $y^3-3y +1$ have?

I graphed the function and it crossed the x-axis $3$ times. But my professor doesn't want a graphical explanation. So in that case, I was looking at the Fundamental Theorem of Algebra and states that a polynomial of degree  n  can have at most  n  distinct real roots. so therefore, there must be 3 real roots? 
EDIT
It seems that there are numerous ways to approach this problem after all. And we can expand this to other types of polynomials as well, not just cubics.

REPLY [2 votes]: This is an approach on how to solve the DEPRESSED cubic equation. To show the roots are real just find the discriminant of the cubic, (it can be shown easily)
To find the roots of your query follow these steps: 
Subsitute $$y= ucosa$$
Substite in the original equation to get
$$(ucosa)^3 -3ucosa+1 =0$$
Recall the identity $$cos(3a)=4cos^3a-3cosa$$
Manipulate the equation to get 
$$u^3(\frac{cos3a +3cosa}{4}) -3ucosa+1 =0$$
Distribute to obtain: 
$$u^3(\frac{cos3a}{4})+ u^3(\frac{3cosa}{4}) -3ucosa+1 =0$$
Take the common factor out of the middle two terms to obtain: 
$$u^3(\frac{cos3a}{4})+ 3ucosa(\frac{u^2}{4}-1)+1 =0$$
"u" is a parameter chosen by us. We are in search of such a parameter "$u$"; so that the middle term cancels out so that we can solve the trigonometric equation. In order for this to occur: $$u=±2$$
Substituting $u=2$ we get
$$2cos3a+1 =0$$
Solving for $cos3a$ we get
$$cos3a= \frac{-1}{2}$$
Therefore 
$$3a= arccos(\frac{-1}{2})$$
$$a = \frac{arccos(\frac{-1}{2})}{3}$$
$$y= 2cos(\frac{arccos(\frac{-1}{2})}{3})$$
Similarly we can subsitute $u=-2$ and find another root. 
Here is a link Wiki
Hope this helped.