Question

If x²yz = 16, xy²z= 81, xyz2 = 256 then x+y+z=​

Answer 1

Answer:

XYZ=256

(2Z)(2Z)Z=256

4Z^3=256

Z^3=64

Z^3=4^3

SO Z=4

AS X=2Z

X=8

Answer 2

Answer:

Add all

x2yz+xy2z+z2yz

=16+81+256

=353

xyz common

xyz(x+y+z)=353

xyz=353/x+y+

then

multiply all

x2yz•y2xz•z2yx

=331776

x4y4z4=331776

xyz=24

xyz(x+y+z)=353

then

24(x+y+z)=353

then

x+y+z=353/