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

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### 3 Answers

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)**

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)**

We notice that the first equation is a difference of cubes and it could be written as:

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

We'll re-write the first equation:

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

We notice that the 2nd factor of the product is the 2nd equation.

But x^2+xy+y^2 = 7

We'll re-write the first equation:

7(x-y) = 7

We'll divide by 7

x - y = 1

x = 1 + y (3)

We'll substitute (3) in (2):

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

We'll expand the square and we'll remove the brackets:

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

We'll combine like terms:

3y^2 + 3y + 1 - 7 = 0

3y^2 + 3y - 6 = 0

We'll divide by 3:

y^2 + y - 2 = 0

y1 = [-1+sqrt(1+8)]/2

y1 = (-1+3)/2

y1 = 1

y2 = (-1-3)/2

y2 = -2

x1 = 1 + 1

x1 = 2

x2 = -1

**The solutions of the system are: {(2 ; 1) ; (-1 ; -2)}.**