If the sum of three numbers is 18 and the sum  of their squares is 36, then find the difference between the sum of their cubes and three times of their product.

Options
a) -1944
b) 1449
c) 4149
d) -1494

Solution

As per question
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

Answer: a) -1944