# 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)

Approved by eNotes Editorial Team We have the system of equations:

x^3 - y^3 = 7

x^2 + xy + y^2 = 7

x^3 - y^3 = 7

=> (x - y)(x^2 + xy + y^2) = 7

=> (x - y) *7 = 7

=> x - y = 1

= x  = 1 + y

x^2 + xy + y^2 = 7

=> (1 + y)^2 + y(1 + y) + y^2 = 7

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

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

=> y^2 + y - 2 = 0

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

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

we get y = 1 and -2

x = 1 + y = 2 and -1

The solution is (2, 1) and (-1, -2)