Question : If $x+y+z=13,x^2+y^2+z^2=133$ and $x^3+y^3+z^3=847$, then the value of $\sqrt[3]{x y z}$ is:
Option 1: $8$
Option 2: $7$
Option 3: $-9$
Option 4: $-6$
Correct Answer: $-6$
Solution :
Given: $x+y+z=13,x^2+y^2+z^2=133$
and $x^3+y^3+z^3=847$
$(x+y+z)^2 = x^2+y^2+z^2+2(xy+yz+zx)$
⇒ $13^2=133+2(xy+yz+zx)$
⇒ $xy+yz+zx=\frac{169-133}{2}=\frac{36}{2}=18$
$x^3+y^3+z^3-3xyz = (x+y+z)(x^2+y^2+z^2-(xy+yz+zx))$
$⇒847-3xyz = 13 ×(133-18)$
$⇒-3xyz = 13×115 - 847 = 1495-847 =648 $
$⇒xyz = \frac{648}{-3} = -216$
$\therefore \sqrt[3]{xyz} = \sqrt[3]{-216} = -6$
Hence, the correct answer is $-6$.
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