What is the sum of the cubes of two numbers if the sum is double their product, that is 6?

First, let's discover the value of the sum of the 2
numbers.

Since it is the double of product, that is 6, that
means that the sum is 12.

Since we know the values of the
sum and the product, we can form the quadratic:

x^2 - 12x +
6 = 0

Now, we'll consider the numbers that verify the
quadratic as x1 and x2:

x1^2 - 12x1 + 6 = 0
(1)

x2^2 - 12x2 + 6 = 0
(2)

We'll add (1) and
(2):

x1^2 + x2^2 = 12(x1 + x2) -
12

But the sum x1 + x2 =
12

x1^2 + x2^2 = 144 - 12

x1^2
+ x2^2 = 132

Now, we'll multiply (1) by
x1:

x1^3 - 12*x1^2 + 6*x1 = 0
(3)

We'll multiply (2) by
x2:

x2^3 - 12*x2^2 + 6*x2 = 0
(4)

We'l add (3) and (4):

x1^3
+ x2^3 = 12(x1^2 + x2^2) - 12*(x1 + x2)

x1^3 + x2^3 =
12*132 - 12*12

x1^3 + x2^3 =
12*(132-12)

x1^3 + x2^3 =
12*120

x1^3 + x2^3 =
1440

The sum of the cubes of the 2 numbers,
whose sum is 12 and product is 6, is: S3 = x1^3 + x2^3 =
1440.