A) \[x-1\]
B) \[{{x}^{2}}+x+1\]
C) \[{{x}^{2}}-x+1\]
D) \[{{x}^{2}}-x-1\]
Correct Answer: B
Solution :
\[{{x}^{3}}-1={{x}^{3}}-{{1}^{3}}\] \[=(x-1)({{x}^{2}}+x+1)\] \[{{x}^{4}}+{{x}^{2}}+1=({{x}^{4}}+2{{x}^{2}}+1)-{{x}^{2}}\] \[={{({{x}^{2}}+1)}^{2}}-{{(x)}^{2}}\] \[=({{x}^{2}}+1+x)({{x}^{2}}+1-x)\] \[=({{x}^{2}}+x+1)({{x}^{2}}-x+1)\] Hence, \[H.C.F.\] of \[{{x}^{3}}-1\] and \[{{x}^{4}}+{{x}^{2}}+1\] is\[{{x}^{2}}+x+1\].
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