a^3+b^3=(a+b)^3-3ab(a+b) determine the sum of the ubes of two numbers if the sum of the two numbers is 4 and the product of the two numbers is 1
\(a^3+b^3=(a+b)^3-3ab(a+b)\) 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
let the two numbers be a and b
\(a+b=4\)
\(ab=1\)
\(a^3+b^3\) to be found.
Well,
\(a^3+b^3=(a+b)^3-3ab(a+b)\)
Substitute
\(a^3+b^3=(4)^3-3(1)(4)\)\(=64-12\)\(=52\)
