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Question: If the cube roots of unity are \(1,\omega ,{\omega ^2}\) , then the roots of the equation \({\left( ...

If the cube roots of unity are 1,ω,ω21,\omega ,{\omega ^2} , then the roots of the equation (x−1)3+8=0{\left( {x - 1} \right)^3} + 8 = 0 are
(a) −1,1+2ω,ω2 - 1,1 + 2\omega ,{\omega ^2}
(b) −1,1−2ω,1−2ω2 - 1,1 - 2\omega ,1 - 2{\omega ^2}
(c) -1, -1, -1
(d) None of these

Explanation

Solution

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,ω,ω21,\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,ω,ω21,\omega ,{\omega ^2} and the equation (x−1)3+8=0{\left( {x - 1} \right)^3} + 8 = 0 has 3 roots.
(x−1)3+8=0{\left( {x - 1} \right)^3} + 8 = 0
Subtract -8 from both the LHS and RHS
(x−1)3+8−8=−8 (x−1)3=−8  {\left( {x - 1} \right)^3} + 8 - 8 = - 8 \\\ {\left( {x - 1} \right)^3} = - 8 \\\
Send the RHS to the LHS
→(x−1)3−8=1 →(x−1)3(−2)3=1 →(x−1−2)3=1  \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
→x−1−2=13\to \dfrac{{x - 1}}{{ - 2}} = \sqrt[3]{1}
Cube roots of unity (1) are 1,ω,ω21,\omega ,{\omega ^2}
∴x−1−2=1\therefore \dfrac{{x - 1}}{{ - 2}} = 1 and x−1−2=ω\dfrac{{x - 1}}{{ - 2}} = \omega and x−1−2=ω2\dfrac{{x - 1}}{{ - 2}} = {\omega ^2}
Solve the above three equations to get the roots of (x−1)3+8=0{\left( {x - 1} \right)^3} + 8 = 0
x−1−2=1 x−1=−2 x=−2+1 x=−1 x−1−2=ω x−1=−2ω x=−2ω+1 x=1−2ω x−1−2=ω2 x−1=−2ω2 x=−2ω2+1 x=1−2ω2  \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 −1,1−2ω,1−2ω2 - 1,1 - 2\omega ,1 - 2{\omega ^2}
The roots of the equation (x−1)3+8=0{\left( {x - 1} \right)^3} + 8 = 0 are −1,1−2ω,1−2ω2 - 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.