A) 1
B) - 1
C) 2
D) - 2
Correct Answer: B
Solution :
Multiplying the numerator and denominator by \[\omega \] and \[{{\omega }^{2}}\] respectively I and II expressions \[=\frac{a+b\omega +c{{\omega }^{2}}}{b+c\omega +a{{\omega }^{2}}}+\frac{a+b\omega +c{{\omega }^{2}}}{c+a\omega +b{{\omega }^{2}}}\] \[=\frac{\omega (a+b\omega +c{{\omega }^{2}})}{(b\omega +c{{\omega }^{2}}+a)}+\frac{{{\omega }^{2}}(a+b\omega +c{{\omega }^{2}})}{(c{{\omega }^{2}}+a+a\omega )}=\omega +{{\omega }^{2}}=-1\]
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