A) 0
B) 1
C) \[{{x}^{2}}+{{y}^{2}}\]
D) \[{{a}^{2}}{{b}^{4}}+{{a}^{2}}{{b}^{2}}\]
Correct Answer: A
Solution :
(a): \[{{a}^{4}}+{{b}^{4}}-{{x}^{2}}{{y}^{2}}=0\] ....... (i) We know, \[{{a}^{6}}+{{b}^{6}}={{\left( {{a}^{2}} \right)}^{3}}+{{\left( {{b}^{2}} \right)}^{3}}=\left( {{a}^{2}}+{{b}^{2}} \right)\left( {{a}^{4}}-{{x}^{2}}{{y}^{2}}+{{b}^{4}} \right)\] \[=\left( {{a}^{2}}+{{b}^{2}} \right)\times 0=0\]
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