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Question -

If a + b + c = 9 and ab + bc + ca = 23, then a3 + b3 + c3 – 3 abc =
(a) 108
(b) 207
(c) 669
(d) 729



Answer -

a3 + b3 + c3 –3abc
= (a + b + c) [a2 + b2 + c2 –(ab + bc + ca)
Now, a + b + c = 9
Squaring,
a2 + b2 + c2 + 2 (ab + be + ca)= 81
⇒ a2 + b2 + c2 + 2 x 23 = 81
⇒ a2 + b2 + c2 + 46 = 81
⇒ a2 + b2 + c2 = 81 – 46 = 35
Now, a3 + b3 + c3 – 3 abc = (a+ b + c) [(a2 + b2 + c2) – (ab + bc+ ca)
= 9[35 -23] = 9 x 12=108                    (a)

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