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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
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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

 

 

 

cool cool cool

CPhill  Oct 12, 2017

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