x1 and x2 are the solutions of x^2+5x-7=0. Prove that x1^3+x2^3 is an integer.

We know that the roots of an equation, substituted in the
equtaion, they verify it.

We'll substitute x1 and x2 in the
original equation:

 x1^2 + 5x1 - 7 = 0
(1)

 x2^2 + 5x2 - 7 = 0
(2)

We'll add (1) and
(2):

(x1^2 +  x2^2) + 5(x1 + x2) - 14 =
0

We'll use Viete's relations to express the sum x1 +
x2:

x1 + x2 = -5

(x1^2
+  x2^2) + 5*(-5) - 14 = 0

(x1^2 +  x2^2) - 25 - 14 =
0

We'll combine like
roots:

(x1^2 +  x2^2) - 39 =
0

We'll add 39 both
sides:

(x1^2 +  x2^2) =
39

Now, we'll re-write the equations (1) and
(2):

 x1^2 + 5x1 - 7 = 0
(1)

 x2^2 + 5x2 - 7 = 0
(2)

We'll multiply (1) by x1 and (2) by
x2:

 x1^3 + 5x1^2 - 7x1 = 0
(3)

 x2^3 + 5x2^2 - 7x2 = 0
(4)

We'll add (3) and
(4):

(x1^3 +  x2^3) + 5(x1^2 +  x2^2) - 7(x1 +  x2) =
0

(x1^3 +  x2^3) = 7(x1 +  x2) - 5(x1^2
+  x2^2)

(x1^3 +  x2^3) = 7*(-5) -
5*(39)

(x1^3 +  x2^3) = -35 -
195

(x1^3 +  x2^3) =
230

It is obvious that the
result of the sum of the cubes is an integer number:
230.