If `(x+ y)^(2) = 21 + z^(2), (y+z)^(2)= 32 + x^(2) and (z+ x)^(2) = 28+ y^(2)`, find `x+ y+ z`=?

Answer

To solve the problem, we have three equations based on the given conditions. Let's break down the steps to find \( x + y + z \).

### Step 1: Rewrite the first equation
We start with the first equation:
\[ (x + y)^2 = 21 + z^2 \]
Rearranging gives:
\[ (x + y)^2 - z^2 = 21 \]
Using the difference of squares identity \( a^2 - b^2 = (a - b)(a + b) \), we can rewrite this as:
\[ (x + y - z)(x + y + z) = 21 \]

### Step 2: Rewrite the second equation
Next, we take the second equation:
\[ (y + z)^2 = 32 + x^2 \]
Rearranging gives:
\[ (y + z)^2 - x^2 = 32 \]
Using the difference of squares identity again, we rewrite this as:
\[ (y + z - x)(y + z + x) = 32 \]

### Step 3: Rewrite the third equation
Now, we move to the third equation:
\[ (z + x)^2 = 28 + y^2 \]
Rearranging gives:
\[ (z + x)^2 - y^2 = 28 \]
Using the difference of squares identity, we rewrite this as:
\[ (z + x - y)(z + x + y) = 28 \]

### Step 4: Summing the equations
Now we have three equations:
1. \( (x + y - z)(x + y + z) = 21 \)
2. \( (y + z - x)(y + z + x) = 32 \)
3. \( (z + x - y)(z + x + y) = 28 \)

Let's denote:
- \( a = x + y + z \)
- \( b = x + y - z \)
- \( c = y + z - x \)
- \( d = z + x - y \)

Now we can express the equations as:
1. \( b \cdot a = 21 \)
2. \( c \cdot a = 32 \)
3. \( d \cdot a = 28 \)

### Step 5: Adding the equations
Adding all three equations gives:
\[ b \cdot a + c \cdot a + d \cdot a = 21 + 32 + 28 \]
This simplifies to:
\[ a(b + c + d) = 81 \]

### Step 6: Finding \( a \)
Now, we can express \( b + c + d \) in terms of \( a \):
\[ b + c + d = (x + y - z) + (y + z - x) + (z + x - y) = (x + y + z) = a \]
Thus, we have:
\[ a \cdot a = 81 \]
This implies:
\[ a^2 = 81 \]
Taking the square root gives:
\[ a = 9 \]

### Final Answer
Thus, we find:
\[ x + y + z = 9 \]