Question

If (x+1) hole square + (y+2) hole square +( z+3) hole square =0 then

Answer 1

Answer:

x+y+z=196

Step-by-step explanation:

(x+1)^2 + (y+2)^2 + (z+3)^2=0

we know that (x+y)^2=x^2+y^2+2xy

x^2+1+2x +y^2+4+4y +z^2+9+6z=0

x^2+y^2+z^2+2(x+2y+3z)=-14

the eq. written at top is a identity

(x+y+z)^2=-14

x+y+z=196

plzzz mark my answer as brainliest