If ab + bc + ca = 8 and a² + b² + c² = 20, then a possible value of $$\frac{1}{2}$$(a + b + c)[(a - b)² + (b - c)² + (c - a)²] is:

A. 72

B. 56

C. 84

D. 80

Answer: Option A

Solution(By Examveda Team)

ab + bc + ca = 8
a² + b² + c² = 20
a² + b² + c² - ab - bc - ca = $$\frac{1}{2}$$[(a - b)² + (b - c)² + (c - a)²]
20 - 8 = $$\frac{1}{2}$$[(a - b)² + (b - c)² + (c - a)²)]
12 = $$\frac{1}{2}$$[(a - b)² + (b - c)² + (c - a)²]
(a + b + c)² = a² + b² + c² + 2(ab + bc + ca)
a + b + c = $$\sqrt {20 + 2 \times 8} = 6$$
$$\frac{1}{2}$$(a + b + c)[(a - b)² + (b - c)² + (c - a)²]
= 12(a + b+ c)
= 12 × 6
= 72