Another way is to cube the second equation: x³ - y³ - 3xy (x - y) = -27 and replace x - y with -3: x³ - y³ = -9 xy - 27 and use this result to substitute x³ - y³ in the first eq.
Let xy=p and solve for p the equation 9p^2+27p+18=0. We'll get p=-2 and p=-1, so xy =-2 and xy=-1. x-y=-3 and xy=-2 gives (x,y)=(-2,1); (-1,2). x-y=-3 and xy=-1 gives (x,y)=((-3+sqrt5)/2; (3-sqrt5)/2); ((-3-sqrt5)/2, (3+sqrt5)/2).
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Nice.
It was unnecessary to restrict yourself to real numbers. It came from the solution.
Another way is to cube the second equation: x³ - y³ - 3xy (x - y) = -27 and replace x - y with -3: x³ - y³ = -9 xy - 27 and use this result to substitute x³ - y³ in the first eq.
Let xy=p and solve for p the equation 9p^2+27p+18=0. We'll get p=-2 and p=-1, so xy =-2 and xy=-1. x-y=-3 and xy=-2 gives (x,y)=(-2,1); (-1,2). x-y=-3 and xy=-1 gives (x,y)=((-3+sqrt5)/2; (3-sqrt5)/2); ((-3-sqrt5)/2, (3+sqrt5)/2).
(x;y)= (-2;1); ( -1; 2);
{(-3+ -√5)/2; ( 3+ -√5)/2}
(y ➖ 3x+2). ➖ 3^1 (y ➖ 3x+1) .
X=-1,-2,((-3+√5)/2), ((-3-√5)/2)
If xy = t, the equations yield t[x^2+y^2+t] =-6 > t[(-3)^2+3t]=-6 > t^2+3t+2=0 > t= -1,-2. If t=-1, y=-1/x > x+1/x=-3 > x = 1/2[-3 +/-√5], y = 1/2[3+/1√5]. If t = -2, x+2/x=-3 > x=-1,-2, y=2,1.
х^3-у^3=(х-у)*((х-у)*2-3ху)
Если ху=а
а*(3-а)=-2
а=-1, а=-2
Ну и х-у=-3, ху=-2 или ху=-1
Решение не составляет труда