Given x+y=3*square root3 and x*y=3 calculate x^2+y^2 and x^4+y^4.

We'll complete the sum of squares, x^2 + y^2, by adding
2xy and creating a perfect square.

x^2 + 2xy +
y^2

Since we've added the amount 2xy, we'll subtract it. In
this way, the result of the sum of squares, will remain
unchanged.

(x^2 + 2xy + y^2 ) - 2xy = (x+y)^2 -
2xy

We'll substitute x + y and xy by their values given in
enunciation:

x^2 + y^2 = (3sqrt3)^2 -
2*3

x^2 + y^2 = 27 - 6

x^2 +
y^2 = 21

Now, we'll create the perfect squares
for:

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

x^4 + y^4 = 21^2 -
2*3^2

x^4 + y^4 = 441 - 18

x^4
+ y^4 = 423

So, the requested sums are: x^2 +
y^2 =  21 and x^4 + y^4 = 423.