Question : If a + b + c = 6 and a2 + b2 + c2 = 38, then what is the value of a(b2 + c2) + b(c2 + a2) + c(a2 + b2) + 3abc?
Option 1: 3
Option 2: 6
Option 3: –6
Option 4: –3
Correct Answer: –6
Solution :
(a + b + c)
2
= a
2
+ b
2
+ c
2
+ 2ab + 2bc + 2ca
⇒ 6
2
= 38 + 2ab + 2bc + 2ca
⇒ (ab + bc + ca) = $\frac{(36 – 38)}{2}$
⇒ (ab + bc + ca) = – 1
Now, a(b
2
+ c
2
) + b(c
2
+ a
2
) + c(a
2
+ b
2
) + 3abc
= a(38 – a
2
) + b(38 – b
2
) + c(38 – c
2
) + 3abc
= 38(a + b + c) – {(a
3
+ b
3
+ c
3
) – 3abc}
= (38 × 6) – {(a + b + c)(a
2
+ b
2
+ c
2
– ab – bc – ca)}
= (38 × 6) – {6 × (38 – (–1))}
= (38 × 6) – 6 × 39
= 6(38 – 39)
= – 6
Hence, the correct answer is –6.
Related Questions
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