If (a + b – 1)2 + (b + c – 9)2 + (c + a – 4)2 ≤ 0, then the value √(a +
b)c + (c +
a)b – 1 is :

(a + b – 1) ²  + (b + c – 9) ²  + (c + a – 4) ² ≤ 0 The value of a, b and c should be 0 or less than 0  for the result to be equal to or less than 0. So, from the given equation: a + b - 1 = 0 a + b = 1      ---- (1) b + c - 9 = 0 b + c = 9      ---- (2) c + a - 4 = 0 c + a = 4       ---- (3) Then, c + a = 4 c + 1 - b = 4       ---- from (1) 9 - b + 1 - b = 4      ---- from (2) - 2b = 4 - 10 b = 3 Substituting the value of b in (1), a + b = 1 a + 3 = 1 a = -2 Substituting the value of b in (2), b + c = 9 3 + c = 9 c = 6 ∴ The value of √ [(a + b)^c + (c + a)^b - 1] is : => √ [((-2) + 3)⁶ + (6 + (-2))³   - 1] ⇒ √ [ (1)⁶ + (4)³ - 1] ⇒ √ [1 + 64 – 1] ⇒ √64 = 8 ∴ The value of √ [(a + b)^c + (c + a)^b - 1] is 8.