Question

Use the identity a^3+b^3=(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

$a^3+b^3=(a+b)^3-3ab(a+b)\\\\a+b=4,\ ab=1\\\\\text{Substitute:}\\\\a^3+b^3=(4)^3-3(1)(4)=64-12=52$