Question : $x,y,$ and $z$ are real numbers. If $x^3+y^3+z^3 = 13, x+y+z = 1$ and $xyz=1$, then what is the value of $xy+yz+zx$?
Option 1: –1
Option 2: 1
Option 3: 3
Option 4: –3
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Correct Answer: –3
Solution :
Given: $x^{3}+y^{3}+z^{3}=13$, $ x+y+z=1$, $xyz=1$
We know, $x^{3}+y^{3}+z^{3}-3xyz= (x^{2}+y^{2}+z^{2}-xy-yz-zx)$
$⇒x^{3}+y^{3}+z^{3}-3xyz= (x+y+z)((x+y+z)^{2}-3(xy+yz+zx))$
Substituting values of $x^{3}+y^{3}+z^{3}$, $ x+y+z$ and $xyz$,
$⇒13-3(1)=(1)((1^{2}-3(xy+yz+zx))$
$⇒xy+yz+zx=\frac{10-1}{-3}$
$\therefore xy+yz+zx=-3$
Hence, the correct answer is –3.
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