By Rational Root Theorem, all rational roots of a polynomial are in the form $\frac{p}{q}$, where $p$ divides the constant term $-8$ and $q$ divides the leading coefficient $1$. List all candidates $\frac{p}{q}$.
$$±8,±4,±2,±1$$
Find one such root by trying out all the integer values, starting from the smallest by absolute value. If no integer roots are found, try out fractions.
$$y=2$$
By Factor theorem, $y-k$ is a factor of the polynomial for each root $k$. Divide $y^{3}-8$ by $y-2$ to get $y^{2}+2y+4$. Solve the equation where the result equals to $0$.
$$y^{2}+2y+4=0$$
All equations of the form $ax^{2}+bx+c=0$ can be solved using the quadratic formula: $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$. Substitute $1$ for $a$, $2$ for $b$, and $4$ for $c$ in the quadratic formula.
$$y=\frac{-2±\sqrt{2^{2}-4\times 1\times 4}}{2}$$
Do the calculations.
$$y=\frac{-2±\sqrt{-12}}{2}$$
Since the square root of a negative number is not defined in the real field, there are no solutions.
