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If a + b + c = 9 and ab + bc + ca = 18, then what is the value of a3 + b3 + c3 – 3abc ?
- A189
- B243
- C361
- D486
Answer not set
Discussion 1
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CCommunity
Since we know that a3 + b3 + c3 – 3abc = (a + b + c)(a2 + b2 + c2 - ab - bc - ca)
And (a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca Using the given data
⇒ 92 = a2 + b2 + c2 + 2(18)
⇒ a2 + b2 + c2 = 45 Put all these things in the required expression
⇒ a3 + b3 + c3 – 3abc = (a + b + c)(a2 + b2 + c2 - ab - bc - ca)
⇒ a3 + b3 + c3 – 3abc = 9 × (45 - 18) = 9 × 27 = 243
∴ a3 + b3 + c3 – 3abc = 243
And (a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca Using the given data
⇒ 92 = a2 + b2 + c2 + 2(18)
⇒ a2 + b2 + c2 = 45 Put all these things in the required expression
⇒ a3 + b3 + c3 – 3abc = (a + b + c)(a2 + b2 + c2 - ab - bc - ca)
⇒ a3 + b3 + c3 – 3abc = 9 × (45 - 18) = 9 × 27 = 243
∴ a3 + b3 + c3 – 3abc = 243