KCET · Maths · Probability
If \( 1, w, w^{2} \) are three cube roots of unity, then \( \left(1-\omega+\omega^{2}\right)\left(1+\omega-\omega^{2}\right) \) is
- A \( 11 \)
- B \( 02 \)
- C \( 03 \)
- D \( 04 \)
Answer & Solution
Correct Answer
(D) \( 04 \)
Step-by-step Solution
Detailed explanation
We have, \(\left(1-\omega+\omega^{2}\right)\left(1+\omega-\omega^{2}\right) \rightarrow(1)\)
We know that, \(1+\omega+\omega^{2}=0\)
\(\Rightarrow 1-\omega+\omega^{2}=-2 \omega\{\) subtracting \(2 \omega\) both the sides \(\}\)
\(\Rightarrow 1+\omega-\omega^{2}=-2 \omega^{2}\left\{\right.\) subtracting \(2 \omega^{2}\) both the sides \(\}\)
So, Eq. (1) becomes
\(=(-2 \omega)\left(-2 \omega^{2}\right)=4 \omega^{3}=4\left\{\right.\) Since,\(\left.\omega^{3}=1\right\}\)
We know that, \(1+\omega+\omega^{2}=0\)
\(\Rightarrow 1-\omega+\omega^{2}=-2 \omega\{\) subtracting \(2 \omega\) both the sides \(\}\)
\(\Rightarrow 1+\omega-\omega^{2}=-2 \omega^{2}\left\{\right.\) subtracting \(2 \omega^{2}\) both the sides \(\}\)
So, Eq. (1) becomes
\(=(-2 \omega)\left(-2 \omega^{2}\right)=4 \omega^{3}=4\left\{\right.\) Since,\(\left.\omega^{3}=1\right\}\)
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