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Find one ordered pair (x,y) of real numbers such that x + y = 6 and x^3 + y^3 = 162 - (x^2 + y^2).

 Jan 1, 2024
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x^3  + y^3  =  162 - (x^2 + y^2)

 

(x + y) (x^2 - xy + y^2)  = 162 - (x^2 + y^2

 

6( x^2 + y^2)  - 6xy = 162 - (x^2 + y^2)

 

7 (x^2 + y^2 + 2xy)  -20xy = 162

 

7 ( x + y)^2  - 20xy = 162

 

7 (6)^2  - 20xy  =162

 

90  = 20xy

 

9/2 = xy

 

(9/2)/x = y

 

x + (9/2)/x  = 6

 

x^2 + (9/2)  = 6x

 

x^2 - 6x   =  -9/2

 

x^2 - 6x + 9  = -9/2 + 9

 

(x - 3)^2  = 9/2

 

x -3  = sqrt(9/2)         or      x - 3  = -sqrt (9/2)

 

x = 3 + 3/sqrt (2)                x =  3 - 3/sqrt (2)

 

 

(x, y)  = ( 3 + 3/sqrt2 , 3- 3/sqrt 2)  and  ( 3 - 3/sqrt2 , 3 + 3/sqrt 2)

 

cool cool cool

 Jan 3, 2024

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