Question : When $a=61, b=63$ and $c=65$, then what is the value of $a^3+b^3+c^3-3abc$?
Option 1: 1456
Option 2: 2268
Option 3: 4536
Option 4: 5460
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Correct Answer: 2268
Solution :
Given:
a = 61, b = 63, and c = 65
$a^3+b^3+c^3-3abc$
= $\frac{(a\:+\:b\:+\:c)[(a\:-\:b)^2\:+\:(b\:-\:c)^2+(c\:-\:a)^2]}{2}$
= $\frac{(61\:+\:63\:+\:65)[(61\:-\:63)^2+(63\:-\:65)^2)+(65\:-\:61)^2]}{2}$
= $\frac{189×24}{2}$
= $2268$
Hence, the correct answer is 2268.
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