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,
$x^4+y^4+z^4+1=4xyz$
$AM\ge GM$
$\therefore \frac{x^4+y^4+z^4+1}{4}\ge {\left(x^4\cdot y^4\cdot z^4\cdot 1\right)}^{1/4}$
$\Rightarrow \frac{4xyz}{4}\ge \left|xyz\right|$
$\Rightarrow xyz>0$
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 $\left(x,y,z\right)=4$