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

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justaguide's profile pic

justaguide | College Teacher | (Level 2) Distinguished Educator

Posted on

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

giorgiana1976's profile pic

giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

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.

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