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

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Let the numbers be A and B.

Their product is 6

=> A*B = 6

Their sum is twice their product

=> A + B = 6*2 = 12

We need the value of A^3 + B^3

(A + B)^3 = A^3 + B^3 + 3*A*B*(A + B)

=> 12^3 = A^3 + B^3 + 3*6*12

=> A^3 + B^3 = 12^3 - 3*6*12

=> A^3 + B^3 = 1728 - 216

=> A^3 + B^3 = 1512

**The sum of the cubes of the two numbers is 1512**

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.**