If the sum of three real numbers is $0$ and their product is $17$, then what is the sum of their cubes?

michaelcai
Oct 12, 2017

#1**+1 **

x + y + z = 0 →- -(x + y) = z → z^3 = - (x^3 + 3x^2y + 3xy^2 + y^3)

xyz = 17

xy [ -(x + y) ] = 17

xy (x + y) = -17 → x^2y + xy^2 = -17 → 3x^2y + 3xy^2 = -51

So......

x^3 + y^3 + [ z^3 ]

x^3 + y^3 + [ - ( x^3 + 3x^2y + 3xy^2 + y^3) ] =

-3x^2y - 3xy^2 =

-[ 3x^2y + 3xy^2] =

- [-51] =

51

CPhill
Oct 12, 2017