Question

Prove that: The sum of cube roots of unity is one.

Answer 1

Let the cube root of $1$ be $x$

$\Rightarrow \sqrt{1}=x$

$\Rightarrow x^3=1$

$\Rightarrow x^3-1=0$

$\Rightarrow (x-1)(x^2+x+1)=0$

$\Rightarrow x-1=0,x^2+x+1=0$

$\Rightarrow x=1,x^2+x+1=0$

$\Rightarrow x=\frac{-1\pm \sqrt{1-4\times 1\times 1}}{2}$ using $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

$\Rightarrow x=\frac{-1\pm \sqrt{-3}}{2}$

$=\frac{-1\pm \sqrt{-1}\sqrt{3}}{2}$

$=\frac{-1\pm i\sqrt{3}}{2}$ where $i=\sqrt{-1}$

$\therefore$ there are $3$ roots $x_2=\frac{-1+i\sqrt{3}}{2},x_3=\frac{-1-i\sqrt{3}}{2},x_1=1$

Hence there are three cubes roots of unity

which are $1,\frac{-1+i\sqrt{3}}{2},\frac{-1-i\sqrt{3}}{2}$

sum of cube roots of unity

$1+\left(\frac{-1+i\sqrt{3}}{2}\right)+\left(\frac{-1-i\sqrt{3}}{2}\right)$

$=1-\frac{2}{2}=0$