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 is 968
answer is 1210
x+y=11 x(Square)+y(Square)=99 121-99=2xy xy=11 x(Cube)+y(Cube)=11(99-11) =11*88 =968
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?
An arithmetic sequence formed of 11 terms, and the sum of all its terms equals to 220. Find the middle term in that sequence.