KCET · Maths · Complex Number
If \(2 x=-1+\sqrt{3} i\), then the value of \(\left(1-x^{2}+x\right)^{6}-\left(1-x+x^{2}\right)^{6}\) is
- A 32
- B 64
- C \(-64\)
- D 0
Answer & Solution
Correct Answer
(D) 0
Step-by-step Solution
Detailed explanation
Given, \(2 x=-1+i \sqrt{3}\)
\(\Rightarrow \quad x=\frac{-1+i \sqrt{3}}{2}=w\)
which is one of the cube root of unity.
\(\therefore \quad\left(1-x^{2}+x\right)^{6}-\left(1-x+x^{2}\right)^{6}\)
\(\begin{aligned}=\left(1-\omega^{2}+\omega\right)^{6}-\left(1-\omega+\omega^{2}\right)^{6} \\(&\left.\because 1+\omega+\omega^{2}=0\right) \end{aligned}\)
\(=\left(-\omega^{2}-\omega^{2}\right)^{6}-(-\omega-\omega)^{6}\)
\(=\left(-2 \omega^{2}\right)^{6}-(-2 \omega)^{6}\)
\(=2^{6}\left\{\omega^{12}-\omega^{6}\right\}\)
\(=2^{6}(1-1) \quad\left(\because \omega^{3}=1\right)\)
\(=2^{6} \cdot 0=0\)
\(\Rightarrow \quad x=\frac{-1+i \sqrt{3}}{2}=w\)
which is one of the cube root of unity.
\(\therefore \quad\left(1-x^{2}+x\right)^{6}-\left(1-x+x^{2}\right)^{6}\)
\(\begin{aligned}=\left(1-\omega^{2}+\omega\right)^{6}-\left(1-\omega+\omega^{2}\right)^{6} \\(&\left.\because 1+\omega+\omega^{2}=0\right) \end{aligned}\)
\(=\left(-\omega^{2}-\omega^{2}\right)^{6}-(-\omega-\omega)^{6}\)
\(=\left(-2 \omega^{2}\right)^{6}-(-2 \omega)^{6}\)
\(=2^{6}\left\{\omega^{12}-\omega^{6}\right\}\)
\(=2^{6}(1-1) \quad\left(\because \omega^{3}=1\right)\)
\(=2^{6} \cdot 0=0\)
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