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Find one ordered pair (x,y) of real numbers such that x + y = 8 and x^3 + y^3 = 200 + x^2 + y^2.

 May 10, 2024
 #1
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x^3 + y^3  = 

 

(x + y) ( x^2 + y^2  - xy) =  (8)  (x^2 + y^2  + 200)

 

This implies that  xy  = -200  →  y = -200/x

 

So

 

x + y  = 8

 

x  - 200/x - 8 = 0

 

x^2 - 8x - 200  = 0

 

x^2 -8x  =  200

 

x^2 - 8x + 16 = 200 + 16

 

(x - 4)^2  = sqrt 216       take both roots

 

x  - 4  = sqrt (216)

 

x = sqrt (216)  + 4          or      x = -sqrt 216  + 4

 

x =  4 + 6sqrt(6)              or      x =    4  -6sqrt (6)

 

y = 4 - 6sqrt (6)                       y  = 4 + 6sqrt (6) 

 

One solution is (x,y)  = ( 4 + 6sqrt(6) , 4 -6sqrt (6) )         

 

 

cool coolcool

 May 10, 2024

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