Question

if [(x-1)^2]+[(y-2)^4]+[(z-3)^6]=0 then the value of xyz is​

Answer 1

Answer:

x=1,,,y=2,,,z=3

Step-by-step explanation:

[(x-1)^2]+[(y-2)^4]+[(z-3)^6]=0

[(1-1)^2]+[(2-2)^4]+[(3-3)^6]=0           ...(x=1,y=2,z=3)

[(0)^2]+[(0)^4]+[(0)^6]=0

0+0+0=0

hence

x,y,z are 1,2,3 respectively....