Solve z^3=-64

$$z ^ { 3 } = - 64$$

Solve for z

$$z=-4$$

Solution Steps

Add 64 to both sides.

$$z^{3}+64=0$$

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.

$$z=-4$$

By Factor theorem, z-k is a factor of the polynomial for each root k. Divide z^{3}+64 by z+4 to get z^{2}-4z+16. Solve the equation where the result equals to 0.

$$z^{2}-4z+16=0$$

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.

$$z=\frac{-\left(-4\right)±\sqrt{\left(-4\right)^{2}-4\times 1\times 16}}{2}$$

Do the calculations.

$$z=\frac{4±\sqrt{-48}}{2}$$

Since the square root of a negative number is not defined in the real field, there are no solutions.

$$z\in \emptyset$$

List all found solutions.

$$z=-4$$