If a^2+b^2+c^2-2a+4b-6c+14=0, find the value of (a-b-c)^2007?

Answer 1

a^2+b^2+c^2-2a+4b-6c+14=0
=>a^2-2a+1+b^2+4b+4+c^2-6c+9=0
=>(a-1)^2+(b+2)^2+(c-3)^2=0
But sum of two or more squares cannot be equal to zero unless each of them is equal to zero
Therefore, (a-1)^2=0 or a-1=0 or a=1
(b+2)^2=0 or b+2=0 or b= -2
(c-3)^2=0 or c-3=0 or c=3
Hence (a-b-c)=1+2-3=0
(a-b-c)^2007=0^2007=0

Answer 2

a^2 + b^2 + c^2 - 2a + 4b - 6c + 14 = a^2 - 2a + 1 -1 +
+ b^2 + 4b + 4 - 4 + c^2 -6c + 9 -9 + 14 =
= (a - 1)^2 + (b + 2)^2 + (c - 3)^2

This is a sum of squares, so a,b,c must be 1, -2 and 3 respectively in order for it to be zero.

a - b - c = 1 - (-2) - 3 = 1 + 2 - 3 = 0

As expected when raising to the 2007th power.