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 their product.

Question

Accepted Answer

Let the three numbers be a, b, and c.
Given:
1. a + b + c = 18
2. a^2 + b^2 + c^2 = 36
We are to find: (a^3 + b^3 + c^3) - 3abc

Recall the identity:
a^3 + b^3 + c^3 = (a + b + c)^3 - 3(a + b + c)(ab + bc + ca) + 3abc
So,
(a^3 + b^3 + c^3) - 3abc = (a + b + c)^3 - 3(a + b + c)(ab + bc + ca)

Let S = a + b + c = 18
Let T = ab + bc + ca
We know:
a^2 + b^2 + c^2 = (a + b + c)^2 - 2(ab + bc + ca)
=> 36 = 18^2 - 2T
=> 36 = 324 - 2T
=> 2T = 324 - 36 = 288
=> T = 144

Now,
(a^3 + b^3 + c^3) - 3abc = S^3 - 3S*T
= 18^3 - 3*18*144
First, calculate 18^3:
18^3 = (18^2)*18 = 324*18 = 5832

Now, 3*18*144 = 54*144 = 7776

So,
(a^3 + b^3 + c^3) - 3abc = 5832 - 7776 = -1944

Therefore, the correct answer is -1944.

Answer

Answer

Answer