Question

If the cube roots of unity are $1,\omega ,{\omega ^2}$ , then the roots of the equation ${\left( {x - 1} \right)^3} + 8 = 0$ are
(a) \[ - 1,1 + 2\omega ,{\omega ^2}\]
(b) \[ - 1,1 - 2\omega ,1 - 2{\omega ^2}\]
(c) -1, -1, -1
(d) None of these

Answer 1

Hint: Given equation is a cubic equation. So there will be 3 roots for this equation. Decompose the given equation into the terms of cube roots of unity $1,\omega ,{\omega ^2}$ , then solve for the roots of the equation.

Complete step-by-step answer:
We are given that the cube roots of unity are $1,\omega ,{\omega ^2}$ and the equation ${\left( {x - 1} \right)^3} + 8 = 0$ has 3 roots.
${\left( {x - 1} \right)^3} + 8 = 0$
Subtract -8 from both the LHS and RHS
$
  {\left( {x - 1} \right)^3} + 8 - 8 = - 8 \\
  {\left( {x - 1} \right)^3} = - 8 \\
 $
Send the RHS to the LHS
$
   \to \dfrac{{{{\left( {x - 1} \right)}^3}}}{{ - 8}} = 1 \\
   \to \dfrac{{{{\left( {x - 1} \right)}^3}}}{{{{\left( { - 2} \right)}^3}}} = 1 \\
   \to {\left( {\dfrac{{x - 1}}{{ - 2}}} \right)^3} = 1 \\
 $
Send the cube power to the RHS
$ \to \dfrac{{x - 1}}{{ - 2}} = \sqrt[3]{1}$
Cube roots of unity (1) are $1,\omega ,{\omega ^2}$
\[\therefore \dfrac{{x - 1}}{{ - 2}} = 1\] and \[\dfrac{{x - 1}}{{ - 2}} = \omega \] and \[\dfrac{{x - 1}}{{ - 2}} = {\omega ^2}\]
Solve the above three equations to get the roots of ${\left( {x - 1} \right)^3} + 8 = 0$
$
  \dfrac{{x - 1}}{{ - 2}} = 1 \\
  x - 1 = - 2 \\
  x = - 2 + 1 \\
  x = - 1 \\
  \dfrac{{x - 1}}{{ - 2}} = \omega \\
  x - 1 = - 2\omega \\
  x = - 2\omega + 1 \\
  x = 1 - 2\omega \\
  \dfrac{{x - 1}}{{ - 2}} = {\omega ^2} \\
  x - 1 = - 2{\omega ^2} \\
  x = - 2{\omega ^2} + 1 \\
  x = 1 - 2{\omega ^2} \\
 $
Therefore, the values of x are \[ - 1,1 - 2\omega ,1 - 2{\omega ^2}\]
The roots of the equation ${\left( {x - 1} \right)^3} + 8 = 0$ are \[ - 1,1 - 2\omega ,1 - 2{\omega ^2}\]
Therefore, from among the options given in the question option B is correct.
So, the correct answer is “Option B”.

Note: A quadratic equation has 2 roots; a cubic equation has 3 roots. So the no. of roots an equation can have is equal to the highest power of x of the equation. When the root values are substituted in the equation, then the result will be zero.