Question

The number of different possible values for the sum $x+y+z$, where $x,y,z$ are real numbers such that $x^4+4y^4+16z^4+64=32xyz$ is

• A. 1
• B. 2
• C. 4
• D. 8

Answer 1

The correct option is C 4

Given :
$x^4+4y^4+16z^4+64=32xyz$....... (i)
From the given equation we can say that $32xyz>0$ .....(ii)
Now applying $A.M\ge G.M$ for the terms,
$x^4,4y^4,16z^4,64$
$\Rightarrow \frac{x^4+4y^4+16z^4+64}{4}\ge (x^4.4y^4.16z^4.64{)}^{\frac{1}{4}}$
using equation (i)
$\Rightarrow 8xyz\ge 8xyz$
$\therefore A.M=G.M$ this condition holds when all the terms are equal,
$x^4=4y^4=16z^4=64x=\pm 2\sqrt{2}y=\pm 2z=\pm \sqrt{2}$
From equation (ii)
Case 1 : when all three are positive
1 way in which all are positive
Case 2 : when any 2 are negative
$(x,y)$
$(z,y)$
$(x,z)$
so 3 ways
Total possible combination of $(x,y,z)$ is 4
Hence number of different possible values for the sum = 4.