A) \[0\]
B) \[1\]
C) \[{{x}^{a+b+c}}\]
D) \[x\]
Correct Answer: B
Solution :
The given expression \[=\frac{{{x}^{3a+3b}}.{{x}^{3b+3c}}.{{x}^{3c+3a}}}{{{x}^{6a+6b+6c}}}\] \[=\frac{{{x}^{3a+3b+3b+3c+3c+3a}}}{{{x}^{6a+6b+6c}}}=\frac{{{x}^{6a+6b+6c}}}{{{x}^{6a+6b+6c}}}\] \[={{x}^{6a+6b+6c-6a-6b-6c}}={{x}^{0}}=1\]You need to login to perform this action.
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