Finding all real solutions in a system of equations?

The system of equations is:
x^3-3xy^2=1 and x3−3xy2=1and

3x^2y-y^3=13x2y−y3=1

1 Answer
George C.
Feb 4, 2018

(x, y) = (1/4root(6)(2)(sqrt(6)+sqrt(2)), 1/4root(6)(2)(sqrt(6)-sqrt(2)))(x,y)=(146√2(√6+√2),146√2(√6−√2))

(x, y) = (-root(3)(4)/2, root(3)(4)/2)(x,y)=(−3√42,3√42)

(x, y) = (-1/4root(6)(2)(sqrt(6)-sqrt(2)), -1/4root(6)(2)(sqrt(6)+sqrt(2)))(x,y)=(−146√2(√6−√2),−146√2(√6+√2))

Explanation:

Given:

{ (x^3-3xy^2=1), (3x^2y-y^3=1) :}

Adding i times the second equation to the first, we get:

1+i = x^3+3ix^2y-3xy^2-iy^3

color(white)(1+i) = x^3+3x^2(iy)+3x(iy)^2+(iy)^3

color(white)(1+i) = (x+iy)^3

Now:

1+i = sqrt(2)(cos (pi/4)+i sin (pi/4))

So:

x+iy = root(6)(2)(cos (pi/12+(2npi)/3) + i sin (pi/12+(2npi)/3))

for n = 0, 1, 2

Equating real and imaginary parts gives us real solutions to the original system:

(x, y) = (root(6)(2)cos(pi/12), root(6)(2)sin(pi/12)) = (1/4root(6)(2)(sqrt(6)+sqrt(2)), 1/4root(6)(2)(sqrt(6)-sqrt(2)))

(x, y) = (root(6)(2)cos((3pi)/4), root(6)(2)sin((3pi)/4)) = (-root(3)(4)/2, root(3)(4)/2)

(x, y) = (root(6)(2)cos((17pi)/12), root(6)(2)sin((17pi)/12)) = (-1/4root(6)(2)(sqrt(6)-sqrt(2)), -1/4root(6)(2)(sqrt(6)+sqrt(2)))

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