Solve for x and y x^3-y^3=7 x^2+xy+y^2=7

x^3 - y^3 =
7..........(1)

x^2 + xy + y^2 = 7
..............(2)

We need to solve the
system.

First we will rewrite equation (1) as a difference
between cubes.

==> x^3 - y^2 = (x-y)(x^2 +
xy+y^2)

But from equation (2) , we know that x^2 + xy+ y^2
= 7

==> 7* (x-y) =
7

Now we will divide by
7.

==> x-y =
1

==> x= y+1
............(3)

Now we will substitute with either
equations.

==> x^3 - y^3 =
7

==> (y+1)^3 - y^3 =
7

==> y^2 + 2y +
1)(y+1)

==> y^3 + 3y^2 +3y + 1 - y^3 =
7

We will reduce
similar.

==> 3y^2 + + 3y -6 =
0

We will divide by
3.

==> y^2 + y -2 =
0

Now we will
factor.

==> (y+2)(y-1) =
0

==> y1= -2 ==> x1= -2+1 =
-1

==> y2= 1 ==> x2= 1+1 =
2

Then we have two solutions:
( -1, -2) and (2, 1)