Counting the Number of Real Roots of $y^{3}-3y+1$ 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.

Just for fun:  we note that since we have a "depressed cubic" of the form $t^3 + pt + q = 0$ (i.e., no quadratic term), the roots have a nice geometric interpretation:
$$
t_k = 2 \sqrt{ -\frac{p}{3}} \cos \left( \frac{1}{3} \arccos \left( \frac{3q}{2p}\sqrt{ - \frac{3}{p} } \right) + \frac{2 \pi k}{3} \right)
$$
for $k = {}$0, 1, or 2.  These can be seen to be the $x$-coordinates of three points equally spaced around the unit circle.  In the given case, we have $p = - 3$ and $q = 1$, so this reduces to
$$
t_k = 2 \cos \left( \frac{1}{3} \arccos\left( -\frac{1}{2} \right) + \frac{2 \pi k}{3} \right) = \left\{ 2 \cos \left( \frac{2\pi}{9} \right), 2 \cos \left( \frac{8\pi}{9} \right), 2 \cos \left( \frac{14\pi}{9} \right) \right\}.
$$
Note that $\cos (14\pi/9) = \cos (4\pi/9)$, so this agrees with Will Jagy's answer as well.
