Question : If $x+y+z=1, \frac{1}{x}+\frac{1}{y}+\frac{1}{z}=1,$ and $xyz=-1$, then $x^3+y^3+z^3 $ is equal to:
Option 1: –1
Option 2: 1
Option 3: –2
Option 4: 2
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Correct Answer: 1
Solution : Given: $x+y+z=1,\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=1$, $xyz=-1$ Consider, $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=1$ ⇒ $xy+yz+zx=xyz=-1$ Now, $x+y+z=1$ ⇒ $(x+y+z)^2=1^2$ ⇒ $x^2+y^2+z^2+2(xy+yz+zx)=1$ ⇒ $x^2+y^2+z^2+2(-1)=1$ ⇒ $x^2+y^2+z^2=3$ We know, $x^3+y^3+z^3-3xyz=(x+y+z)(x^2+y^2+z^2-(xy+yz+zx)$ So, putting all the values, we get, ⇒ $x^3+y^3+z^3-3×(-1)=1×[3-(-1)]$ ⇒ $x^3+y^3+z^3=4-3$ $\therefore x^3+y^3+z^3=1$ Hence, the correct answer is 1.
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