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If a + b + c = 15 and 1 + 1 + 1 = 71 ,then the value of a3 + b3 + c3 – 3abc is a b c abc
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- 160
- 180
- 200
- 220
Correct Option: B
a + b + c = 15,
| + | + | = | ||||
| a | b | c | abc |
| ⇒ | = | ||
| abc | abc |
⇒ ab + bc + ca = 71
∴ a3 + b3 + c3 – 3abc = (a + b + c) (a2 + b2 + c2 – ab – bc – ac)
= (a + b + c) {(a + b + c)2 – 3 (ab + bc + ac)}
= 15 (152 – 3 × 71)
= 15 (225 – 213) = 15 × 12
= 180
