x1^3 + x2^3 = 8

x^2 - x -c = 0

We know thatL

x1 + x2 = -b/a = 1

x1*x2 = c/a = c

f(x1) = x1^2 -x1 -c = 0

Multiply by x1:

==> x1^3 - x1^2 -cx1 = 0.........(1)

f(x2) = x2^2 -x2-c = 0

Multiply by x2:

==> x2^3 - x2^2 -cx2 = 0..........(2)

add (1) and (2)

==> x1^3 + x2^= x1^2 + x2^2 + cx1 + cx2

==> 8 = (x1+x2)^2 - 2x1x2 + c(x1+c2)

==> 8 = 1 - 2(-c) + c(1)

==> 8 = 1+2c+c

==> 3c= 7

==> c= 7/3

x^2-x-c = 0. given x1^3+x2^3 = 8.

Solution:

Let x^2+bx+k = 0 and x1 andx2 are roots.

By the reation between the roots and coeffients,

x1+x2 = -b/a.

x1x2 = k/a

Here, a=1, b =-1 and k = -c

x1+x2 = - (-1/1) = 1

x1x2 = -c/1 = -c

given x1^3+x2^3 = 8.

We use the identity,

x1^3+x2^3 = (x1+x2)^3 - 3x1*x2 (x1+x2)

8 = (1)^3 - 3(-c)(1).

8-1 = 3c

c = 7/3

In order to determine the coefficient c, we'll apply Viete's relationships.

We'll put the solutions, x1 and x2, into the equation.

x1^2-x1-c=0 (1)

x2^2-x2-c=0 (2)

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

x1^3-x1^2-c*x1=0 (3)

x2^3-x2^2-c*x2=0 (4)

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

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

But, from enunciation, x1^3+x2^3=8 and

(x1^2+x2^2)=(x1+x2)^2-2x1*x2

x1+x2=1

x1*x2=-c

(x1^2+x2^2)=(1)^2+2*c

We'll substitute the values into (5):

(8) - (1+2c) - c*(1)=0

We'll combine like terms:

-8-1-2c-c=0

-3c=9

We'll divide by -3 both sides:

**c=-3**