If a + b + c = 11 and ab + bc + ca = 28, then find the value of a3 + b3 + c3 - 3abc.
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Correct answer: B
Explanation
The correct answer is option 2 i.e. 407
We know, a3 + b3 + c3 - 3abc = (a + b + c)[(a + b + c)2 - 3(ab + bc + ca)]
⇒ 11 × [(11)2 - 3(28)]
⇒ 11 × [121 - 84]
⇒ 11 × 37 = 407