If the value of `(a + b - 3)^(2) + (b + c- 2)^(2) + (c + a - 5)^(2) = 0`, then the value of `sqrt((b+c)^(a)+(c+a)^(b))` is :

Options

A. 3
B. 1
C. 0
D. 2

Answer

To solve the equation

\[ (a + b - 3)^{2} + (b + c - 2)^{2} + (c + a - 5)^{2} = 0, \]

we need to analyze the terms in the equation. Since the sum of squares is zero, each individual square must also be zero. Therefore, we can set up the following equations:

1. \( a + b - 3 = 0 \)
2. \( b + c - 2 = 0 \)
3. \( c + a - 5 = 0 \)

Now, let's solve these equations step by step.

### Step 1: Solve for \( a + b \)

From the first equation:

\[ a + b = 3 \tag{1} \]

### Step 2: Solve for \( b + c \)

From the second equation:

\[ b + c = 2 \tag{2} \]

### Step 3: Solve for \( c + a \)

From the third equation:

\[ c + a = 5 \tag{3} \]

### Step 4: Substitute to find individual variables

Now we can express \( c \) in terms of \( a \) using equation (3):

\[ c = 5 - a \tag{4} \]

Substituting equation (4) into equation (2):

\[ b + (5 - a) = 2 \]
\[ b + 5 - a = 2 \]
\[ b - a = -3 \tag{5} \]

Now we can express \( b \) in terms of \( a \):

\[ b = a - 3 \tag{6} \]

### Step 5: Substitute \( b \) back into equation (1)

Substituting equation (6) into equation (1):

\[ a + (a - 3) = 3 \]
\[ 2a - 3 = 3 \]
\[ 2a = 6 \]
\[ a = 3 \]

### Step 6: Find \( b \) and \( c \)

Now substituting \( a = 3 \) back into equation (6):

\[ b = 3 - 3 = 0 \]

Now substituting \( b = 0 \) into equation (2):

\[ 0 + c = 2 \]
\[ c = 2 \]

### Step 7: Summary of values

We have found:

- \( a = 3 \)
- \( b = 0 \)
- \( c = 2 \)

### Step 8: Calculate \( \sqrt{(b+c)^{a} + (c+a)^{b}} \)

Now we need to calculate:

\[ \sqrt{(b+c)^{a} + (c+a)^{b}} = \sqrt{(0 + 2)^{3} + (2 + 3)^{0}} \]

Calculating each term:

- \( (0 + 2)^{3} = 2^{3} = 8 \)
- \( (2 + 3)^{0} = 5^{0} = 1 \)

Now substituting back:

\[ \sqrt{8 + 1} = \sqrt{9} = 3 \]

### Final Answer

The value of \( \sqrt{(b+c)^{a} + (c+a)^{b}} \) is \( 3 \).

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