Question

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

Answer 1

Step-by-step explanation:

${(a - 1)}^{2} + {(b + 2)}^{2} + {(c + 1)}^{2} = 0$

${a}^{2} - 2a + 1 + {b}^{2} + 4b + 4 + {c}^{2} + 1 - 2c = 0$

${a}^{2} + {b}^{2} + {c}^{2} - 2a + 4b - 2c + 6 = 0$