COMEDK · Maths · 3. Complex Number
If the cube roots of unity are \(1, \omega, \omega^{2}\), then the roots of the equation \((x-1)^{3}+8=0\) are
- A \(-1,-1,-1\)
- B \(-1,-1+2 \omega .-1-2 \omega^{2}\)
- C \(-1,1+2 \omega, 1+2 \omega^{2}\)
- D \(-1,1-2 \dot{w} \mathbf{i}-2 \dot{w}^{2}\)
Answer & Solution
Correct Answer
(D) \(-1,1-2 \dot{w} \mathbf{i}-2 \dot{w}^{2}\)
Step-by-step Solution
Detailed explanation
We have, \(\begin{aligned} & &(x-1)^{3} &=-8=(-2)^{3} \\ \therefore & & x-1 &=-2,-2 \omega,-2 \omega^{2} \\ \Rightarrow & & x &=-1,1-2 \omega, 1-2 \omega^{2} \end{aligned}\)
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