A) \[g(1)=h(1)=0\]
B) \[g(1)=h(1)\ne 0\]
C) \[g(1)-h(1)=1\]
D) \[g(1)+h(1)=1\]
Correct Answer: A
Solution :
We have, \[{{x}^{2}}+x+1=(x-\omega )(x-{{\omega }^{2}})\] Now,\[p(x)=g({{x}^{3}})+xh({{x}^{3}})\]is divisible by \[{{x}^{2}}+x+1.\] \[\Rightarrow \] \[x=\omega \] and\[x={{\omega }^{2}}\]are roots of\[p(x)=0\]. \[\Rightarrow \] \[p(\omega )=0,p({{\omega }^{2}})=0\] \[\Rightarrow \] \[g(1)+\omega h(1)=0\] and \[g(1)+{{\omega }^{2}}h(1)=0\] \[\therefore \] \[g(1)=0=h(1)\]
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