If $^{'} n^{'}$ Odd integer, then $(1 + i)^{6 n} + (1 - i)^{6 n}$ is equal to :

0

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Solution

The correct option is A $0$
$\begin{matrix} N o w, {(1 + i)}^{6 n} + {(1 - i)}^{6 n} = {({(1 + i)}^{2})}^{3 n} + {({(1 - i)}^{2})}^{3 n} = {(2 i)}^{3 n} + {(- 2 i)}^{3 n} i f n i s o d d i i n t e r g e r a b o v e r e s s i o n i s Z e r o . O p t i o n A i s c o r r e c t a n s w e r . \end{matrix}$

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