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The number of triples $(x, y, z)$ of real numbers satisfying the equation $x^{4}+y^{4}+z^{4}+1=4 x y z$ is
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Verified Answer
The correct answer is:
4
$(x, y, z)$ are real & $x^{4}, y^{4}, z^{4}$ are positive real numbers
$$
\begin{array}{l}
\therefore \frac{x^{4}+y^{4}+z^{4}+1}{4} \geq|x y z| \\
\Rightarrow(x y z) \geq|x y z| \\
\text { i.e., } x y z>0
\end{array}
$$
So it holds equality
$\therefore \mathrm{x}^{4}=\mathrm{y}^{4}=\mathrm{z}^{4}=1 ;$ But $\mathrm{xyz}>0$
$\therefore(\mathrm{x}, \mathrm{y}, \mathrm{z}) \in\{(1,1,1),(1,-1,-1),(-1,1,-1),(-1,-1,1)\}$
So no. of triplets is 4 .
$$
\begin{array}{l}
\therefore \frac{x^{4}+y^{4}+z^{4}+1}{4} \geq|x y z| \\
\Rightarrow(x y z) \geq|x y z| \\
\text { i.e., } x y z>0
\end{array}
$$
So it holds equality
$\therefore \mathrm{x}^{4}=\mathrm{y}^{4}=\mathrm{z}^{4}=1 ;$ But $\mathrm{xyz}>0$
$\therefore(\mathrm{x}, \mathrm{y}, \mathrm{z}) \in\{(1,1,1),(1,-1,-1),(-1,1,-1),(-1,-1,1)\}$
So no. of triplets is 4 .
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