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I'm not sure how to deal with the cube roots.

 

\[\large\begin{eqnarray}\begin{split}x+y=&3383\\\sqrt[3]x+\sqrt[3]y=&17\\xy=&?\end{split}\end{eqnarray}\]

 Feb 2, 2021
 #1
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x^(1/3)  + y^(1/3)   =  17       square both sides

(x^1/3  + y^1/3)^2   = 17^2

(x^2/3  + 2(xy)^1/3  + y^2/3)  =  289

x^2/3  + y^2/3  =  289   - 2(xy)^ (1/3)

 

Factor x + y as a sum of  cubes

 

x + y    =     (x^1/3  + y^1/3)  ( x^(2/3)  - (xy)^1/3 + y^(2/3)  )

 

3383  =(17)  ( x^(2/3)  - (xy)^(1/3)  + y^(2/3)  )

 

3383  /17    = x^(2/3) + y^(2/3)  - (xy)^(1/3) 

 

199  =  289  - 2(xy)^(1/3 )  - (xy)^(1/3)

 

189 - 289   =  - 3(xy)^(1/3)

 

-90  = -3(xy)^(1/3)

 

-90  / -3  =  (xy)^1/3

 

30  =  (xy)^(1/3      cube both sides

 

27000   =xy

 

 

cool cool cool

 Feb 2, 2021
edited by CPhill  Feb 2, 2021

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