Question 1105564
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(x+y)(x^2+y^2)=40    (1)
(x-y)(x^2-y^2)=16    (2)

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x^3 + xy^2 + yx^2 + y^3 = 40   (1')
x^3 - xy^2 - yx^2 + y^3 = 16   (2')


Add (1') and {2') (both sides). You will get

2(x^3 + y^3) = 56,     or  x^3 + y^3 = 28.     (3)


Subtract (2') from (1'). You will get

2(xy^2 + yx^2) = 24,   or  xy^2 + x^2y = 12.   (4)


Then

x^3 + 3x^2y + 3xy^2 + y^3 = 28 + 3*12 = 64,   or

(x+y)^3 = 64,   which implies   

x + y = 4       (5)


Next, from (4) you have

xy*(x+y) = 12,  and,  replacing here x+y by 4 (due to (5)), you get

xy*4 = 12,   or

xy = {{{12/4}}} = 3.    (6)


From (5) and (6) you can easily get the 


<U>Answer</U>.  (x,y) = (1,3)   or  (x,y) = (3,1).
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Solved.