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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).

 May 27, 2024
 #1
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x + y =  6     square both sides

 

x^2 + 2xy + y^2 = 36

x^2 + y^2  = 36 -2xy

 

x^3 + y^3  = 162 - (x^2 + y^2)

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

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

 6 (36  - 2xy -xy) = 162 - (36 -2xy)

216 - 18xy = 126 + 2xy

90 = 20xy

xy = 9/2

y = 9/(2x)

 

x + y = 6

x + 9/(2x) = 6

2x^2 + 9 = 12x

2x^2 -12x + 9 = 0

 

x =  [12 + sqrt [ 144 - 72] ] / 4  =  [ 12 + 6sqrt2] / 4  =  3 + 3/sqrt 2

 

y =  3 - 3/sqrt 2

 

 

cool cool cool

 May 28, 2024

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