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=32 xyz` is (A) 1 (B) 2 (C) 4 (D) 8

Options

A. 1
B. 2
C. 4
D. 8

Answer

C

Explanation

Applying Am`ge`gm.
`(x^(4)+4y^(4)+16z^(4)+64)/(4) ge(4^(6)x^(4)y^(4)z^(4))^(1//4)`
`x^(4)+4y^(4)+16z^(4)+64ge32|xyz|`
So equal when each term is equal
`:. X^(4)+4y^(4)=16z^(4)=64`
`impliesx=pm2sqrt(2)`
`y=pm2`
`x= pmsqrt(2)`
For x.y.z
For `X^(4)+4y^(4)+16z^(4)_64=32xyz`
Either each of x,y,z is (+)ve`to` 1 case/
Or two of x,y,z are (-)ve `to ` 3 cases
`:.` 4 cases of different (x,y,z) triplets
`:.` 4 possible x+y+z values (as `x ne y ne z`)