Let $f(x)=x^3-15x+1$; then $f\,'(x)=3x^2-15=3(x^2-5)=3(x-\sqrt5)(x+\sqrt5$. It’s easy to check that $f$ has a local maximum at $x=-\sqrt5$ and a local minimum at $x=\sqrt5$, either by the second derivative test, by examining the sign of the first derivative, or simply by knowing the shape of the graph of a cubic polynomial.
so $f(x)$ changes sign three times on the interval $[-4,4]$ and by the intermediate value theorem must have (at least) three zeroes.
Note that there are many ways to choose the test points, but using the critical points is an easy way to be sure of hitting good ones.
Added: To show that there are only three solutions to the original equation, apply Rolle’s theorem to $f$ to show that between any two zeroes of $f$ there must be a critical point.