TS EAMCET · Maths · Complex Number
If \(\omega\) is a complex cube root of unity, then \((x+1)(x+\omega)(x-\omega-1)\) is equal to
- A \(x^3-1\)
- B \(x^3+1\)
- C \(x^3+2\)
- D \(x^3-2\)
Answer & Solution
Correct Answer
(A) \(x^3-1\)
Step-by-step Solution
Detailed explanation
\(\omega \rightarrow\) cube root of unity ie, \(\quad \omega^3=1\) and \(1+\omega+\omega^2=0\) Then, \((x+1)(x+\omega)(x-\omega-1)\) \(=(x+1)\left(x^2+\omega x-\omega x-\omega^2-x-\omega\right)\) \(=(x+1)\left(x^2-x-\left(\omega+\omega^2\right)\right)\)…
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