Question

The number of triples $(x, y, z)$ of real number satisfying the equation $x^{4} + y ^{4} + z^{4} + 1 = 4xyz$ is

• A. 0
• B. 4
• C. 8
• D. more than 8

Answer 1

Answer: (B)

We have,
$x^{4}+y^{4}+z^{4}+1=4xyz$
$AM \ge\,GM$
$\therefore \frac{x^{4}+y^{4}+z^{4}+1}{4} \ge \left(x^{4}\cdot y^{4}\cdot z^{4}\cdot1\right)^{1/ 4}$
$\Rightarrow \frac{4xyz}{4} \ge\,\left|xyz\right|$
$\Rightarrow xyz >\,0 $
It is possible $\left(1, 1, 1\right)\left(-1, -1, 1\right)\left(-1, 1, -1\right) , \left(1, -1, 1\right)$
So number of triplet $\left(x, y,z\right)=4$