A) \[0\]
B) \[-1\]
C) \[{{a}^{2}}+{{b}^{2}}\]
D) \[{{a}^{2}}{{b}^{4}}+{{a}^{4}}{{b}^{2}}\]
Correct Answer: A
Solution :
[a] \[\because \] \[{{A}^{3}}+{{B}^{2}}=(A+B)({{A}^{2}}+{{B}^{2}}-AB)\] On putting \[A={{a}^{2}}\]and \[B={{b}^{2}},\]we get\[({{a}^{6}}+{{b}^{6}})\] \[=({{a}^{2}}+{{b}^{2}})({{a}^{4}}+{{b}^{4}}-{{a}^{2}}{{b}^{2}})\] \[=({{a}^{2}}+{{b}^{2}})({{a}^{2}}{{b}^{2}}-{{a}^{2}}{{b}^{2}})\] \[=({{a}^{2}}+{{b}^{2}})\times 0=0\] |
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