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If a = 34, b = c = 33, then the value of a3 + b3 + c3 – 3abc is
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- 0
- 111
- 50
- 100
- 0
Correct Option: D
Using Rule 22,
| a3 + b3 + c3 - 3abc = | (a + b + c)[ (a - b)2 + (b - c)2 + (c - a)2 ] | |
| 2 |
Here , a = 34, b = c = 33
| ⇒ a3 + b3 + c3 - 3abc = | (34 + 33 + 33)[ (34 - 33)2 + (33 - 33)2 + (33 - 34)2 ] | |
| 2 |
| ⇒ a3 + b3 + c3 - 3abc = | × 100[ 1 + 0 + 1 ] = 100 | |
| 2 |
