Solve a^3=-1

$$a ^ { 3 } = - 1$$

Solve for a

$$a=-1$$

Solution Steps

Add 1 to both sides.

$$a^{3}+1=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term 1 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±1$$

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.

$$a=-1$$

By Factor theorem, a-k is a factor of the polynomial for each root k. Divide a^{3}+1 by a+1 to get a^{2}-a+1. Solve the equation where the result equals to 0.

$$a^{2}-a+1=0$$

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, -1 for b, and 1 for c in the quadratic formula.

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

Do the calculations.

$$a=\frac{1±\sqrt{-3}}{2}$$

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

$$a\in \emptyset$$

List all found solutions.

$$a=-1$$