You should also use the following alternative method that focuses more on special products.

Notice that you may obtain the sum of powers `x^4 + y^4` if you raise to square `x^2+y^2` such that:

`x^4 + y^4= (x^2 + y^2)^2 - 2x^2*y^2`

Notice that the first equation states that `x^2 + y^2 = 0` , hence `x^4 + y^4 = - 2x^2*y^2` .

Notice that the second equation states that`x^4 + y^4 = 8` , hence - 2x^2*y^2 = 8.

`x^2*y^2 = -4`

By Lagrange's resolvents, you may form the monic equation `z^2 - pz + q=0 ` substituting 0 for p and -4 for q such that:

`z^2 - 0*z - 4 = 0`

`z^2 - 4 = 0 =gt (z-2)(x+2) = 0`

`z- 2 = 0 =gt z_1 = 0`

`z+ 2 = 0 =gtz_2 = -2`

You need to solve the equations `x^2 = 2` and `y^2 = -2` such that:

`x_(1,2) = +-sqrt2`

`y_(1,2) = +-sqrt(-2)` (from complex number theory,`sqrt(-1) = +-i)` `y_(1,2) = +-i*sqrt2`

**Hence, evaluating all the solutions to the symmetrical system of equations yields the eight possible combinations: `(sqrt2,i*sqrt2) ; (-sqrt2,-i*sqrt2) ;(-sqrt2,i*sqrt2) ; (sqrt2,-i*sqrt2);(i*sqrt2,sqrt2) ; (-i*sqrt2,-sqrt2); (-i*sqrt2,sqrt2) ; (i*sqrt2,-sqrt2). ` **

Solve the system `x^2+y^2=0,x^4+y^4=8` .

`x^2+y^2=0==>y^2=-x^2==>y=+-sqrt(-x^2)==>y=+-ix`

`x^4+y^4=8==>y^4=8-x^4` Then

`y^2=+-sqrt(8-x^4)==>y=+-iroot(4)(8-x^4)`

Then there are 8 possible equations; solving one of them:

`ix=iroot(4)(8-x^4)`

`(ix)^4=[iroot(4)(8-x^4)]^4`

`x^4=8-x^4` (Since `i^4=1` )

`2x^4=8`

`x^4=4`

`x^2=+-2,+-2i`

`x=+-sqrt(2),+-isqrt(2)`

** Note that all eight equations will have the same solutions. `(ix)^4=(-ix)^4` and `[root(4)(8-x^4)]^4=[-root(4)(8-x^4)]^4=[iroot(4)(8-x^4)]^4=[-iroot(4)(8-x^4)]^4`

Plugging the values for x into `y=+-ix` yields the solutions.

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**The eight solutions are:**

`(sqrt(2),isqrt(2)),(sqrt(2),-isqrt(2))`

`(-sqrt(2),isqrt(2)),(-sqrt(2),-isqrt(2))`

`(isqrt(2),sqrt(2)),(isqrt(2),-sqrt(2))`

`(-isqrt(2),sqrt(2)),(-isqrt(2),-sqrt(2))`

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