Solving for \(a + b + c\)

The problem provides the equation: \(a^2 + b^2 + c^2 + 3 = 2(a + b + c)\) We need to find the value of \((a + b + c)\).

Algebraic Manipulation

First, rearrange the equation to bring all terms to one side:

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

Group the terms involving \(a\), \(b\), and \(c\) separately:

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

Completing the Square

To solve this, we use the technique of completing the square for each variable. The general form is \((x - k)^2 = x^2 - 2kx + k^2\). We need \(k^2\) terms.

• For \((a^2 - 2a)\), we need \(+1\).
• For \((b^2 - 2b)\), we need \(+1\).
• For \((c^2 - 2c)\), we need \(+1\).

We can rewrite the equation by adding and subtracting these required terms:

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

Combine the perfect square terms and the constants:

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

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

This simplifies to:

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

Finding the Values of a, b, and c

The sum of squares of real numbers can only be zero if each individual term is zero. Therefore:

• \((a - 1)^2 = 0 \implies a - 1 = 0 \implies a = 1\)
• \((b - 1)^2 = 0 \implies b - 1 = 0 \implies b = 1\)
• \((c - 1)^2 = 0 \implies c - 1 = 0 \implies c = 1\)

Calculating the Final Value

Now, calculate the value of \((a + b + c)\):

\((a + b + c) = 1 + 1 + 1 = 3\)

The value of \((a + b + c)\) is 3.