Question

Prove that the sum of cube roots of unity is zero.

Answer 1

Hint: Write the cubic equation representing the cube root of unity and then solve it to find the three solutions. Then add them to get the sum of the cube roots of unity.

Complete step-by-step answer:
The cube roots of unity are defined as the numbers which when raised to the power of 3 give the result as 1. It is the cube root of the number 1.
It is represented by the Greek letter $$\omega$$.
Now, let us solve to get the cube roots of unity.
$$\omega =\sqrt[3]{1}$$
Taking cubes on both sides of the equation, we get:
$${\omega }^3=1$$
Taking 1 to the left-hand side of the equation, we get:
$${\omega }^3-1=0.........(1)$$
Using the formula $$a^3-b^3=(a-b)(a^2+ab+b^2)$$, we get the following:
$${\omega }^3-1^3=(\omega -1)({\omega }^2+\omega +1^2)$$
Simplifying, we get:
$${\omega }^3-1=(\omega -1)({\omega }^2+\omega +1)........(2)$$
Using equation (2) in equation (1), we get:
 $$(\omega -1)({\omega }^2+\omega +1)=0$$
Hence, we get:
$$\omega -1=0;{\omega }^2+\omega +1=0$$
Hence, one of the solutions is $$\omega =1$$.
We find the other two solutions using the equation $${\omega }^2+\omega +1=0$$.
The roots of the quadratic equation $$a{x}^2+bx+c=0$$ are given as follows:
$$x={\displaystyle \frac{-b\pm \sqrt{b^2-4ac}}{2a}}$$
The roots of the equation $${\omega }^2+\omega +1=0$$ is then given as follows:
$$\omega ={\displaystyle \frac{-1\pm \sqrt{1^2-4(1)(1)}}{2(1)}}$$
$$\omega ={\displaystyle \frac{-1\pm \sqrt{1-4}}{2}}$$
$$\omega ={\displaystyle \frac{-1\pm \sqrt{-3}}{2}}$$
We know that $$\sqrt{-3}$$ is a complex number and can be written as $$i\sqrt{3}$$
$$\omega ={\displaystyle \frac{-1\pm i\sqrt{3}}{2}}$$
Hence, the three cube roots of unity are $${\displaystyle \frac{-1+i\sqrt{3}}{2}}$$, $${\displaystyle \frac{-1-i\sqrt{3}}{2}}$$, and 1.
Adding the cube roots of unity, we get as follows:
$${\displaystyle \frac{-1+i\sqrt{3}}{2}}+{\displaystyle \frac{-1-i\sqrt{3}}{2}}+1=-{\displaystyle \frac{1}{2}}-{\displaystyle \frac{1}{2}}+1$$
Simplifying, we get:
$${\displaystyle \frac{-1+i\sqrt{3}}{2}}+{\displaystyle \frac{-1-i\sqrt{3}}{2}}+1=-1+1$$
$${\displaystyle \frac{-1+i\sqrt{3}}{2}}+{\displaystyle \frac{-1-i\sqrt{3}}{2}}+1=0$$
Hence, we proved that the sum of the cube roots of unity is zero.

Note: You can also use the sum of roots of the cubic equation $$a{x}^3+b{x}^2+cx+d=0$$ which is $${\displaystyle \frac{b}{a}}$$ to get the sum of the cube roots of unity.