Question : If $\frac{x}{y}+\frac{y}{x}=1$ and $x+y=2$, then the value of $x^3+y^3$ is:
Option 1: 1
Option 2: 3
Option 3: 0
Option 4: 2
Correct Answer: 0
Solution : $\frac{x}{y}+\frac{y}{x}=1$ and $x+y=2$ $⇒\frac{x^2+y^2}{xy}=1$ $⇒{x^2+y^2}=xy$ ___(i) $x^3+y^3=(x+y)(x^2+y^2-xy)$ From (i), $⇒x^3+y^3=0$ Hence, the correct answer is 0.
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Question : If $\frac{1}{x+2}=\frac{3}{y+3}=\frac{1331}{z+1331}=\frac{1}{3}$, then what is the value of $\frac{x}{x+1}+\frac{y}{y+6}+\frac{z}{z+2662}$?
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Question : If $x^2 = y+z$, $y^2=z+x$, $z^2=x+y$, then the value of $\frac{1}{x+1}+\frac{1}{y+1}+\frac{1}{z+1}$ is:
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