Question

If the cube roots of unity are 1 ,$\omega$ , ${\omega }^2$ then the roots of the equation ${(x-1)}^3$ + 8 = 0 are

• A. -1 , 1+ 2 , 1 +
• B. -1 , 1-2 , 1-
• C. -1 , -1 , -1
• D. -1 , -1 + 2 , - 1-

Answer 1

The correct option is B

-1 , 1-2 , 1-

When we are finding the cube root of -8 , we will three roots. First one is easy to guess, which is -2

We get the other two roots when we proceed the following way.
${(x-1)}^3$ = -8
(x - 1) = -2 ${(+1)}^{\frac{1}{3}}$ (we will get three roots when we find the cube roots af 1 / unity)
${\left(1\right)}^{\frac{1}{3}}$ is cube root of unity
They are 1 ,$\omega$ , ${\omega }^2$
$\Rightarrow$x = 1 - 2${\left(1\right)}^{\left(\frac{1}{3}\right)}$
$\Rightarrow$ x= 1 - 2 , 1- 2$\omega$ , 1 - $2{\omega }^2$
= -1 , 1- 2$\omega$ , 1 - ${2\omega }^2$