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
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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