A) \[1\]
B) \[-2\]
C) \[0\]
D) \[-4\]
Correct Answer: D
Solution :
\[{{(1+i)}^{3}}+{{(1-i)}^{3}}={{(1+i)}^{2}}(1+i)+{{(1-i)}^{2}}(1-i)\] \[=(1+{{i}^{2}}+2i)(1+i)+(1+{{i}^{2}}-2i)\,(1-i)\] \[\left[ \begin{align} & \because \,\,\,\,\,{{(a+b)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab \\ & {{(a-b)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab \\ \end{align} \right]\] \[=2i(1+i)+(-2i)(1-i)\] \[(\because \,\,{{i}^{2}}=-1)\] \[=2i+2{{i}^{2}}-2i+2{{i}^{2}}\] \[=4{{i}^{2}}=-4\] \[(\because \,\,{{i}^{2}}=-1)\]
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