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 = a2 + b2 + c2 + 2ab + 2bc + 2ca ⇒ 62 = 38 + 2ab + 2bc + 2ca ⇒ (ab + bc + ca) = $\frac{(36 – 38)}{2}$ ⇒ (ab + bc + ca) = – 1 Now, a(b2 + c2) + b(c2 + a2) + c(a2 + b2) + 3abc = a(38 – a2) + b(38 – b2) + c(38 – c2) + 3abc = 38(a + b + c) – {(a3 + b3 + c3) – 3abc} = (38 × 6) – {(a + b + c)(a2 + b2 + c2 – ab – bc – ca)} = (38 × 6) – {6 × (38 – (–1))} = (38 × 6) – 6 × 39 = 6(38 – 39) = – 6 Hence, the correct answer is –6.
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Question : If a3 + b3 = 217 and a + b = 7, then the value of ab is:
Option 1: – 6
Option 2: – 1
Option 3: 7
Option 4: 6
Question : If a + b + c = 19, ab + bc + ca = 120, then what is the value of a3 + b3 + c3 – 3abc?
Option 1: 18
Option 2: 23
Option 3: 31
Option 4: 19
Question : If x6 – 512y6 = (x2 + Ay2) (x4 – Bx2 y2 + Cy4), then what is the value of (A + B – C)?
Option 1: –80
Option 2: –72
Option 3: 72
Option 4: 48
Question : If a + b + c = 5 and ab + bc + ca = 7, then the value of a3 + b3 + c3 – 3abc is:
Option 1: 15
Option 2: 20
Option 3: 25
Option 4: 30
Question : If $a=331, b=336$ and $c=–667$, then the value of $a^3+b^3+c^3–3abc$ is:
Option 1: 1
Option 2: 63
Option 3: 3
Option 4: 0
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