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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:
Answer & Solution
Correct Answer:
Option
A
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
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
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