Determine all possible pairs(x,y) of real
numbers that satisfy this system of equations:
(x+y)(x^2+y^2) = 85
(x-y)(x^2-y^2) = 45
Expanded out the equations are
x^3+x^2*y+x*y^2+y^3 = 85
x^3-x^2*y-x*y^2+y^3 = 45
Multiply the first equation by 2 and subtract the second, then simplify:
x^3+3*x^2*y+3*x*y^2+y^3 = 125.
Both sides are perfect cubes, since we are over reals we can just take the principal cube root.
Then x+y=5.
Go back to the second equation and finish factoring:
(x-y)^2*(x+y)=45
Substitute x+y=5, divide out the 5, and take square roots to get two branches:
x-y=3 or x-y=-3.
Then solving the two possible linear systems gets two solutions: (x,y) = (4,1) or (1,4).
Solution