If the value of $(a+b-2)^2+(b+c-5)^2+(c+a-5)^2=0$, then the value of $\sqrt{(b+c)^a+(c+a)^b-1}$ is:
Answer & explanation
Correct answer: option 3
Given,
(a + b - 2)2 + (b + c - 5)2 + (c + a - 5)2 = 0
So,
(a + b - 2) = 0
= a + b = 2 ---- (1)
(b + c - 5) = 0
= b + c = 5 ---- (2)
(c + a - 5) = 0
= c + a = 5 ---- (3)
Solving, (1), (2) and (3), we get
a = 1, b = 1 and c = 4
Then, $\sqrt{(b+c)^a+(c+a)^b-1}$ = $\sqrt{(1+4)^1+(4+1)^1-1}$ = 3