RRB NTPC · Question 11
If a3 + b3 + c3 – 3abc = 0, then find the value of (a2/bc + b2/ac – 3). A. –c2/ab B. –c2/bc C. –c3/ba D. –c/a
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Answer: (C) A
Explanation
a³ + b³ + c³ − 3abc = 0 gives a + b + c = 0 (the usual case), so a³ + b³ + c³ = 3abc and a³ + b³ = 3abc − c³. Then a²/bc + b²/ac − 3 = (a³ + b³)/abc − 3 = (3abc − c³)/abc − 3 = −c³/abc = −c²/ab.