A) 0
B) 1
C) -1
D) 2
Correct Answer: A
Solution :
\[\frac{a}{b}+\frac{b}{a}=1\Rightarrow \frac{{{a}^{2}}+{{b}^{2}}}{ab}=1\] \[\Rightarrow \] \[{{a}^{2}}+{{b}^{2}}=ab\] \[\Rightarrow \]\[{{a}^{2}}+{{b}^{2}}-ab=0\] \[\therefore \]\[{{a}^{3}}+{{b}^{3}}\] \[=(a+b)({{a}^{2}}-ab+{{b}^{2}})=0\]You need to login to perform this action.
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