Question

If $\omega$ is a complex cube root of unity , find the equation , whose roots are 2 , 2$\omega$ , 2${\omega }^2$

• A. - + x -8 = 0
• B. + 8 =0
• C. - 8 + 8x -8 = 0
• D. - 8 = 0

Answer 1

The correct option is D

- 8 = 0

We will solve this in two methods.

$\underset{–}{Methods1}$

The equation whose roots are 1,$\omega$ and ${\omega }^2$ is $x^3$ - 1 =0
To get the equation whose roots are 2, 2$\omega$ and 2${\omega }^2$, we will replace x by $\frac{x}{2}$ in the equation $x^3$ -1 = 0. This is because the when the roots are multiplied by 2 , we have divide x by 2 (It is covered in detail in quadratic / theory of equations)
$\Rightarrow$ The equation whose roots are 2 , 2$\omega$ , 2${\omega }^2$ is ${\left(\frac{x}{2}\right)}^3$ - 1 = 0
or $x^3$ - 8 =0

$\underset{–}{Methods2}$

In this method we will find the equation by multiplying all the factors.
The required equation is
(x - 2)(x - 2$\omega$)(x - 2${\omega }^2$)
= $x^3$ - (2 + 2$\omega$ + 2${\omega }^2$)$x^2$ + (2 × 2$\omega$ + 2$\omega$ $\times$ 2${\omega }^2$ + 2 $\times$ 2${\omega }^2)x$ - (2 $\times$ 2$\omega$ $\times$ 2${\omega }^2$)
= $x^3$ - 2( 1 + $\omega$ + ${\omega }^2$) $x^2$ +4(2 + 1 +${\omega }^2)x$ -8
= $x^3$ - 8