Finding all real solutions in a system of equations?

The system of equations is:
#x^3-3xy^2=1 and #

#3x^2y-y^3=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) = (-root(3)(4)/2, root(3)(4)/2)#

#(x, y) = (-1/4root(6)(2)(sqrt(6)-sqrt(2)), -1/4root(6)(2)(sqrt(6)+sqrt(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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