A) \[(2a+b+1)(4{{a}^{2}}+{{b}^{2}}+1-2ab-b-a)\]
B) \[(2a+b+1)(4{{a}^{2}}+{{a}^{2}}+1-2abb-2a)\]
C) \[(2a+b+1)(4{{a}^{2}}+{{b}^{2}}+1-2ab-b-2a)\]
D) None of the above
Correct Answer: C
Solution :
\[8{{a}^{3}}+{{b}^{3}}-6ab+1\] \[={{(2a)}^{3}}+{{b}^{3}}+{{1}^{3}}-3(2a)(b)(1)\] \[=(2a+b+1)(4{{a}^{2}}+{{b}^{2}}+1-2ab-b-2a)\]
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