Question

if the ordered triplets of real no. (x,y,z) satisfy √x-y+z=√x-√y+√z,x+y+z=8 and x-y+z=4 ,then find the value of xyz​

Answer 1

Answer:

For some reason, all the existing answers so far assume that x,y and z must be positive.

I’ll provide a more general answer. First, add all equations together, to get 2(x+y+z)2=288 (several answers did the same thing).

This means, (x+y+z)=±12 .

First, let’s consider the positive solution:

x+y+z=12 . This means that x+y=10 , which immediately implies z=2 (since z=(x+y+z)−(x+y)=12−10 ). Likewise, from the other equations, we get x=4,y=6 .

Now for the negative solution:

x+y+z=−12 , implying x+y=−10 , so we get z=−2 , and likewise x=−4 , y=−6 .

So, the full analytical solution consists of the two solutions (−4,−6,−2),(4,6,2)