Solve the simultaneous equations x^3+y^3=-7 and x^2+y^2=5?

We'll apply the formula of the sum of the
cubes:

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

We'll replace the sum x+y by S and the product x*y by
P.

The system will become:

S(5
- P) = -7 (1)

S^2 - 2P = 5

S^2
= 5 + 2P (2)

We'll raise to square
(1):

S^2*(5-P)^2 = 49

(5 +
2P)(5-P)^2 = 49

We'll raise to square the
binomial:

(5 + 2P)(25-10P + P^2) =
49

125 - 50P + 5P^2 + 50P - 20P^2 + 2P^3 =
49

We'll eliminate like
terms:

-15P^2 + 2P^3 + 76 =
0

We notice that one root is P =
-2

-60 - 16 + 76 = 0

We'll
re-write 2P^3 - 15P^2 + 76 = (P+2)(2P^2 - 19P + 38)

We'll
cancel 2P^3 - 15P^2 + 76 = 0

(P+2)(2P^2 - 19P + 38) =
0

2P^2 - 19P + 38 = 0

P1 =
[19+sqrt(361 - 304)]/4

P1 = (19+sqrt
57)/4

P2 = (19 - sqrt
57)/4

Since S(5 - P) = -7 then we'll
have:

P = -2 => S(5+2)=-7 => S =
-1

P = (19+sqrt 57)/4 => S =
(1+sqrt57)/2

P = (19-sqrt 57)/4 => S =
(1-sqrt57)/2

Now, we'll solve the
systems:

1) x + y = -1 and xy =
-2

x^2 + x - 2 = 0

x = -2 and
y = 1 or x = 1 and y = -2

2) x + y = (1+sqrt57)/2 and x*y =
(19+sqrt 57)/4

x^2 - (1+sqrt57)*x/2 + (19+sqrt 57)/4 =
0

delta = (1+sqrt57)^2 - 4*19-4*sqrt
57

delta = 1 + 2sqrt 57 + 57 - 76 -
4*sqrt57

delta = -18 - 2sqrt 57 <
0

Since the discriminant is negative, this system has no
solutions.

Since the system is symmetrical,
the solutions of the system of equations are represented by the pairs: (1 , -2) and (-2
, 1).