Given:
The sum of the three numbers is 18
a+b+c = 18
The sum of their squares is 36
a2+b2+c2 = 36
We need to find the difference between the sum of their cubes and three times their product
a3+b3+c3−3abc
There is a well-known identity for the sum of cubes of three numbers
a3+b3+c3−3abc=(a+b+c)(a2+b2+c2−ab−bc−ca)
(a+b+c)2 = a2+b2+c2+2(ab+bc+ca)
182 = 36 + 2(ab+bc+ca)
2(ab+bc+ca) = 324 - 36
2(ab+bc+ca) = 288
ab+bc+ca = 144
a3+b3+c3−3abc=(a+b+c)(a2+b2+c2−ab−bc−ca)
a3+b3+c3−3abc= (18) x (36 - 144)
a3+b3+c3−3abc= 18 x 108
a3+b3+c3−3abc= -1944
Option (b) is the correct answer.
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