The value of $a^3 + b^3 + c^3 - 3abc$, when a = 125, b = 127 and c = 129, is :
Answer & explanation
Correct answer: option 2
a = 125
b = 127
c = 129
We know that,
a3 + b3 + c3 - 3abc = \(\frac{1}{2}\) × (a + b + c) × [(a - b)2 + (b - c)2 + (c - a)2]
According to the question,
= a3 + b3 + c3 - 3abc = \(\frac{1}{2}\) × (125 + 127 + 129) × [(125 - 127)2 + (127 - 129)2 + (125 - 129)2]
= a3 + b3 + c3 - 3abc = \(\frac{1}{2}\) × 381 × [4 + 4 + 16]
= a3 + b3 + c3 - 3abc =\(\frac{1}{2}\) × 381 × 24
= a3 + b3 + c3 - 3abc = 12 × 381
= a3 + b3 + c3 - 3abc = 4572