Solve k^3=1

$$k ^ { 3 } = 1$$

Solve for k

$$k=1$$

Solution Steps

Subtract 1 from both sides.

$$k^{3}-1=0$$

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.

$$k=1$$

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

$$k^{2}+k+1=0$$

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.

$$k=\frac{-1±\sqrt{1^{2}-4\times 1\times 1}}{2}$$

Do the calculations.

$$k=\frac{-1±\sqrt{-3}}{2}$$

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

$$k\in \emptyset$$

List all found solutions.

$$k=1$$