We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive pseudonymised information about your use of our website. cookie policy and privacy policy.
 
+0  
 
0
591
1
avatar+601 

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

 Oct 12, 2017
 #1
avatar+102422 
+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

 

 

 

cool cool cool

 Oct 12, 2017

17 Online Users