A) 0
B) 1
C) 2
D) \[{{(2)}^{a+b+c}}\]
Correct Answer: B
Solution :
[b] \[{{\left( \frac{{{2}^{a}}}{{{2}^{b}}} \right)}^{a+b}}{{\left( \frac{{{2}^{b}}}{{{2}^{c}}} \right)}^{b+c}}{{\left( \frac{{{2}^{c}}}{{{2}^{a}}} \right)}^{c+a}}\] \[={{({{2}^{a-b}})}^{a+b}}.{{({{2}^{b-c}})}^{b+c}}.{{({{2}^{c-a}})}^{c+a}}\] \[=2({{a}^{2}}-{{b}^{2}})+({{b}^{2}}-{{c}^{2}})+({{c}^{2}}-{{a}^{2}})={{2}^{0}}=1\]You need to login to perform this action.
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