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

=================>

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 =  = 3.    (6)


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


Answer.  (x,y) = (1,3)   or  (x,y) = (3,1).

He's no more help than the back of the book. LOL
She didn't give the complex solutions.



Notice that if we interchange x and y, the system of equations
simplifies to the same system, so if (x,y)=(p,q) is a solution,
then so is (x,y)=(q,p) 

Let y = kx







Solve each for x³



Equate the expressions for x³



Divide both sides by 8



Cross-multiply:



Factor 1-k² as the difference of squares:



Get 0 on the right



Factor out common factor (1+k)











Use the zero factor property:

1+k = 0;    -3k²+10k-3 = 0
  k = -1;    3k²-10k+3 = 0
           (3k-1)(k-3) = 0  
         3k-1 = 0;  k-3 = 0
           3k = 1;    k = 3
            k = 1/3

We substitute those three values for k in:



Substituting k=-1



, not possible:

Substituting k=1/3


 





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Use zero-factor property:

x=3; x²+3x+9=0
     
     
     
     
     
     

And since y = kx = (1/3)x

when x = 3, y = (1/3)(1) = 1

So one solution is (x,y) = (3,1), and by swapping x and y, we
have

 (x,y) = (1,3)

and when ,  

So two more solutions are



and



And two more solutions by swapping x and y are



and



Edwin