A) 1
B) -1
C) \[\frac{1}{2}\]
D) 0
Correct Answer: D
Solution :
\[\frac{x}{y}+\frac{y}{x}=1\] \[\Rightarrow \,\,\,\,\,\,{{x}^{2}}+{{y}^{2}}=xy\Rightarrow {{x}^{2}}-xy+{{y}^{2}}=0\] Now, \[{{x}^{3}}+{{y}^{3}}=\left( x+y \right)\left( {{x}^{2}}-xy+{{y}^{2}} \right)\] = \[\left( x+y \right)\times 0=\mathbf{0}\]You need to login to perform this action.
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