Question

Use the identity
a3+b3=(a+b)3−3ab(a+b)a3+b3=(a+b)3−3ab(a+b) to determine the sum of the cubes of two numbers if the sum of the two numbers is 4 and the product of the two numbers is
1

Answer 1

The sum of the cubes of two numbers is 52

Explanation:

Let a and b be two numbers.

It is given that the sum of the two numbers is 4 and the product of the two numbers is 1

Thus, Rewriting it in the expression form, we have,

$a+b=4$

    $ab=1$

To determine the sum of the cubes of two numbers, let us substitute these values in the identity $a^3+b^3=(a+b)^3-3ab(a+b)$ , we get,

$a^3+b^3=(4)^3-3(1)(4)$

           $=64-12$

           $=52$

Thus, the sum of the cubes of two numbers is 52