A) 8
B) 0
C) 2
D) None of these
Correct Answer: A
Solution :
[a] \[\left| \begin{matrix} a & b & c \\ b & c & a \\ c & a & b \\ \end{matrix} \right|=0\] \[\Rightarrow 3acb-{{a}^{3}}-{{b}^{3}}-{{c}^{3}}=0\] \[\Rightarrow {{a}^{3}}+{{b}^{3}}+{{c}^{3}}=3abc\] \[{{2}^{\frac{{{a}^{2}}}{bc}}}{{.2}^{\frac{{{b}^{2}}}{ca}}}{{.2}^{\frac{{{c}^{2}}}{ab}}}={{2}^{\frac{{{a}^{3}}+{{b}^{3}}+{{c}^{3}}}{abc}}}={{2}^{3}}=8\]You need to login to perform this action.
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