Question

If $1 , \omega , \omega^2$ are the cube roots of unity, then $\frac{1}{1+2\omega} + \frac{1}{2+\omega } - \frac{1}{1+\omega } = $

• A. 1
• B. $\omega$
• C. $\omega^2$
• D. 0

Answer 1

The Correct Option is D

Cube roots of unity mean \(\omega^3=1\) and \(1+\omega+\omega^2 = 0 →\)

\(\frac{1}{1+2\omega}+\frac{1}{2+\omega}-\frac{1}{1+\omega}\)

\(=\frac{(2+\omega)+(1+2\omega)}{(1+2\omega)(2+\omega)}-\frac{1}{1+\omega}\)

\(=\frac{3+3\omega}{\omega+2\omega^{2}+2+4\omega}-\frac{1}{1+\omega}\)

\(=\frac{3(1+\omega)}{2\omega^{2}+5\omega+2}-\frac{1}{1+\omega}\)

\(=\frac{(3+3\omega)(1+\omega)-(2\omega^2+5\omega+2}{(2\omega^{2}+5\omega+2)(1+\omega)}\)

\(=\frac{3+3\omega+3\omega+3\omega^2-2\omega^2-5\omega-2}{2\omega^2+5\omega+2+2\omega^3+5\omega^2+2\omega}\)

\(=\frac{\omega^2+\omega+1}{2\omega^3+7\omega^2+7\omega+2}\)

\(=\frac{0}{2\omega^{3}+7\omega^{2}+7\omega+2}\)

\(=0\)