SOLUTION: (1)/(x^(3))+(1)/(y^(3))=35, (1)/(x^(2))-(1)/(xy)+(1)/(y^(2))=7 How do you solve these two equations in a system?

Introduce new variables  a = ,  b = .


Then the given system can be re-written in this EQUIVALENT form


    a^3 + b^3 = 35        (1)     (from the first equation)

and

    a^2 - ab + b^2 = 7    (2)     (from the second equation).



From Algebra, we know this identity

    a^3 + b^3 = (a+b)*(a^2 - ab + b^2).


Therefore, equation (1) is

    (a+b)*(a^2 - ab + b^2) = 35.    (3)


In equation (3), replace a^2 - ab + b^2 by 7, based on (2).   You will get then

    7*(a+b) = 35                    


from (3),  which implies

    a + b = 35/7 = 5.


So, instead of two equations (1) and (2), one of which is of the degree 3 and another is of the degree 2,
we get an EQUIVALEN system of equations

    a^3 + b^3 = 35      (4)    (the same as equation (1) )

    a   + b   =  5      (5)    (deduced and has the degree 1)


Now, from equation (5)  express  b = 5-a  and substitute it into equation (4).  You will get then

    a^3 + (5-a)^3 = 35

    a^3 + 5^3 - 3*5^2*a + 3*5*a^2 - a^3 = 35

    15a^2 - 75a + 125 = 35

    15a^2 - 75a + 90 = 0

      a^2 - 5a  +  6 = 0      (after canceling factor 15 in previous equation)

      (a - 2)*(a - 3) = 0      (after factoring the previous equation).


So the system (1), (2)  has two solutions

    (a)  (a,b) = (2,3)   

and

    (b)  (a,b) = (3,2).


It means that the original system has two solutions

    (a)  x = ,  y = 

and

    (b) x = ,  y = .


ANSWER.  The given system has two solutions  (a)  x = ,  y =   and  (b) x = ,  y = .