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
Correct Answer: 19
Solution : Given, a + b + c = 19 and ab + bc + ca = 120 We know, a 3 + b 3 + c 3 – 3abc = (a + b + c) [(a + b + c) 2 – 3(ab + bc + ca)] = 19[19 2 – 3 × 120] = 19[361 – 360] = 19 × 1 = 19 Hence, the correct answer is 19.
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