$$x^{3}+81=17$$
Subtract $17$ from both sides.$$x^{3}+81-17=0$$
Subtract $17$ from $81$ to get $64$.$$x^{3}+64=0$$
By Rational Root Theorem, all rational roots of a polynomial are in the form $\frac{p}{q}$, where $p$ divides the constant term $64$ and $q$ divides the leading coefficient $1$. List all candidates $\frac{p}{q}$.$$±64,±32,±16,±8,±4,±2,±1$$
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.$$x=-4$$
By Factor theorem, $x-k$ is a factor of the polynomial for each root $k$. Divide $x^{3}+64$ by $x+4$ to get $x^{2}-4x+16$. Solve the equation where the result equals to $0$.$$x^{2}-4x+16=0$$
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 $a$, $-4$ for $b$, and $16$ for $c$ in the quadratic formula.$$x=\frac{-\left(-4\right)±\sqrt{\left(-4\right)^{2}-4\times 1\times 16}}{2}$$
Do the calculations.$$x=\frac{4±\sqrt{-48}}{2}$$
Since the square root of a negative number is not defined in the real field, there are no solutions.$$x\in \emptyset$$
List all found solutions.$$x=-4$$