<h2>How to find the value of (a-b)² + (b-c)² + (c-a)² When a + b + c = 6 and a² + b² + c² = 14?</h2>

Problem Statement:

If a + b + c = 6 and a² + b² + c² = 14, find the value of (a-b)² + (b-c)² + (c-a)².

Solution:

The formula for (a-b)² + (b-c)² + (c-a)² is:

(a-b)² + (b-c)² + (c-a)² = 2(a² + b² + c²) - 2(ab + bc + ca).

From the problem:

a + b + c = 6 → (a + b + c)² = 36.

Expand (a + b + c)²:

a² + b² + c² + 2(ab + bc + ca) = 36.

Substitute a² + b² + c² = 14:

14 + 2(ab + bc + ca) = 36.

2(ab + bc + ca) = 22 → ab + bc + ca = 11.

Now substitute values into the formula:

(a-b)² + (b-c)² + (c-a)² = 2(a² + b² + c²) - 2(ab + bc + ca).

= 2(14) - 2(11) = 28 - 22 = 6.

Final Answer:

The value is 6.