If the sum of two numbers is 11, and the sum of their squares is 99, find the sum of their cubes.

Question

The sum of 3 single digit numbers is 15 less than their product . If we subtract 2 from first given number then sum of these numbers will become 7 more than their product. Then what will be the product of given 3 numbers?

Answer 1

a+b = 11
a^2+b^2=99
=>
(a+b)^2 = a^2+b^2+2ab
121=99+2ab
or
ab=11

Now lets look at
(a+b)(a^2+b^2) = a^3+ab(a+b)+b^3
11*99 = a^3+b^3+11*11
a^3+b^3 = 968

Answer 2

Answer is 968

Answer 3

answer is 1210

Answer 4

x+y=11
x(Square)+y(Square)=99
121-99=2xy
xy=11
x(Cube)+y(Cube)=11(99-11)
=11*88
=968