solve the system for x and y?

Question

4x^3 + 4y = 0

and

4y^3 + 4x = 0

Answer 1

4x^3 + 4y = 0 (1)
4y^3 + 4x = 0 (2)

(1)&(2) : divide both sides by 4
x^3 + y = 0 (1')
y^3 + x = 0 (2')

y = -x^3 subsitute into (2')
(-x^3)^3 + x = 0
-x^9 + x = 0
-x^8 + 1 = 0
x^8 = 1
x = +/- 1

At x = +1; y = -1
At x = -1; y = 1

Answer 2

4x^3 + 4y = 0
4y^3 + 4x = 0
x^3 + y = 0
y^3 + x = 0
y = - x^3
(- x^3)^3 + x = 0
- x^9 + x = 0
x^9 - x = 0
x(x^8 - 1) = 0
x = 0,
x^8 = 1
x = 0, 1, (1/2)(√2 + i√2), i, - (1/2)(√2 - i√2), - 1,
- (1/2)(√2 + i√2), - i, (1/2)(√2 - i√2)

Answer 3

fr [1] y = -x^3
subst to [2]
4 (-x^3)^3 + 4x = 0
-x^9 + x = 0
x(-x^8 +1)= 0
x(1 + x^4)(1 + x^2)(1 + x)(1 - x) = 0
x=0; with y = -x^3 = 0
x^4= -1 , with 4 complex roots
x^2 = -1, with 2 complex roots
x = -1 ; with y = -x^3 = 1
x = 1 ; with y = -(1)^3 = -1

Answer 4

4x^3 + 4y = 0
4y^3 + 4x = 0

4x^3=- 4y
4y^3 =- 4x

x^3=- y
y^3 =- x


y=-x^3

(-x^3)^3=-x

-x^9=-x

x(x^8-1)=0

x=0 y=0

x^8=1

x=1
y=-1