A) 1
B) 2
C) 0
D) 3
Correct Answer: C
Solution :
\[{{a}^{3}}-{{b}^{3}}-{{c}^{3}}=0,\] \[\therefore \,\,\,{{a}^{3}}-{{b}^{3}}={{c}^{3}}\] Taking cube of both side: \[{{({{a}^{3}}-{{b}^{3}})}^{3}}={{({{c}^{3}})}^{3}}\Rightarrow {{a}^{9}}-{{b}^{9}}-3{{a}^{3}}{{b}^{3}}({{a}^{3}}-{{b}^{3}})={{c}^{9}}\] \[{{a}^{9}}-{{b}^{9}}-3{{a}^{3}}{{b}^{3}}{{c}^{3}}={{c}^{9}}\] \[{{a}^{9}}-{{b}^{9}}-{{c}^{9}}-3{{a}^{3}}{{b}^{3}}{{c}^{3}}=\underline{\mathbf{0}}\]You need to login to perform this action.
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