Of course there is :-) A technique for solving cubics has been known since the 16th century.
Let's start with $x^3+ax+b=0$, which is your case. We'll see later that it's always possible to eliminate the $x^2$ term, so this is in fact the only case we really need to worry about.
The trick is to make the problem initially look harder by setting $x=p+q$. Instead of a single variable we now have two; we hope to use the extra freedom to gain something, and we shall.
Substituting, we get
$(p+q)^3+a(p+q)+b=0$
or
$p^3+3p^2q+3pq^2+q^3+a(p+q)+b=0$
and now we observe that $3p^2q+3pq^2$ can be rearranged:
$p^3+q^3+3pq(p+q)+a(p+q)+b=0$
and finally, we observe the two terms with a common $p+q$:
$p^3+q^3+(3pq+a)(p+q)=0$.
This seems about as bad as where we started - however, we have this extra degree of freedom: we can force $3pq+a$ to be 0. If we do that, by setting $q = -\frac{a}{3p}$, we get
$p^3 - \frac{a^3}{27p^3} + b = 0$
and we're basically done: renaming $p^3$ as $z$, this is a quadratic equation in $z$. Solve it, determine $p$, calculate $q$, and the solution to the original equation is $x=p+q$ (of course there could be multiple solutions).
All of this can be recast as explicit equations for the solution, but I never remember those; the only way I can remember this is by recalling the $x=p+q$ trick.
Finally, if you have
$x^3+ax^2+bx+c=0$
you can make a simple linear change of variables $x=y+r$, whereby this becomes
$y^3+(3r+a)y^2+\ldots=0$
so choosing $r = -a/3$ makes the $y^2$ term vanish. Solve for $y$ as before and remember the shift $x=y-a/3$ to get the solution for the original equation.