Question : If $x+y+z=17, x y z=171$ and $x y+y z+z x=111$, then the value of $\sqrt[3]{\left(x^3+y^3+z^3+x y z\right)}$ is:
Option 1: –64
Option 2: 4
Option 3: 0
Option 4: –4
Correct Answer: –4
Solution : Given: $x + y + z = 17,xy + yz + zx = 111$ and $xyz = 171$ $(x + y + z)^2 = x^2 + y^2 + z^2 + 2(xy + yz + zx)$ $⇒17^2 = x^2 + y^2 + z^2 + 2 × 111$ $⇒x^2 + y^2 + z^2 = 289 - 222$ $⇒x^2 + y^2 + z^2 = 67$ Using algebraic identity $x^3 + y^3 + z^3 - 3xyz = (x + y + z)[x^2 + y^2 + z^2 – (xy + yz + zx)]$ $⇒x^3 + y^3 + z^3 - 3 × 171 = 17 × (67 - 111)$ $⇒x^3 + y^3 + z^3 - 513 = -748$ $⇒x^3 + y^3 + z^3 = -748 + 513$ $⇒x^3 + y^3 + z^3 = -235$ So, $\sqrt[3]{(x^3+y^3+z^3+xyz)} = \sqrt[3]{(-235+171)}=-4$ Hence, the correct answer is –4.
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