Prove the following

Question:

Let $A=\left[\begin{array}{lll}x & y & z \\ y & z & x \\ z & x & y\end{array}\right]$, where $x, y$ and $z$ are real numbers such that

$x+y+z^{2}>0$ and $x y z=2$. If $A^{2}=I_{3}$, then the value of $x^{3}+y^{3}+z^{3}$ is

Solution:

$\mathrm{A}=\left[\begin{array}{ccc}\mathrm{x} & \mathrm{y} & \mathrm{z} \\ \mathrm{y} & \mathrm{z} & \mathrm{x} \\ \mathrm{z} & \mathrm{x} & \mathrm{y}\end{array}\right] \quad \therefore|\mathrm{A}|=\left(\mathrm{x}^{3}+\mathrm{y}^{3}+\mathrm{z}^{3}-3 \mathrm{xyz}\right)$

$\mathrm{A}^{2}=\mathrm{I}_{3}$

$\left|A^{2}\right|=1$

$\therefore\left(\mathrm{x}^{3}+\mathrm{y}^{3}+\mathrm{z}^{3}-3 \mathrm{xyz}\right)^{2}=1$

$\Rightarrow \mathrm{x}^{3}+\mathrm{y}^{3}+\mathrm{z}^{3}-3 \mathrm{xyz}=1$

$\Rightarrow \mathrm{x}^{3}+\mathrm{y}^{3}+\mathrm{z}^{3}=6+1=7$

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