Three real numbers $x,y,z$ are such that $x^{2} + 6y = - 17,y^{2} + 4z = 1$ and $z^{2} + 2x = 2$. Th

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Three real numbers $x,y,z$ are such that $x^{2} + 6y = - 17,y^{2} + 4z = 1$ and $z^{2} + 2x = 2$. Then the value of $x^{3} + y^{3} + z^{3}$ is equal to

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Adding we get$$\begin{matrix} & & \ \ \ \ \ \ \ \ & x^{2} + 6y + y^{2} + 4z + z^{2} + 2x & \ = - 14 \ \Rightarrow & & \ \ \ \ \ \ \ \ & \ (x + 1)^{2} + (y + 3)^{2} + (z + 2)^{2} & \ = 0 \ & \ \Rightarrow & \ \ \ \ \ \ \ \ & x & \ = - 1,y = - 3,z = - 2 \end{matrix}$$

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