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.