Which is the sum of the cubes of the roots of equation x^2+ax-4=0?

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f(x)= x^2+ax-4

Assume that the roots are x1 and x2

then:

x1^2+ax1-4=0 ....(1)

x2^2+ax2-4=0 ....(2)

add (1) and (2)

==> x1^2+x2^2+ax1+ax2-8=0

==> x1^2+x2^2= 8-a(x1+x2)

but (x1+x2)= -a

==> x1^2+x2^2=8+a^2......(3)

Now we need to calculate the sum of the cubes of the roots:

that means we need x1^3+x2^3

Let us multiply...

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f(x)= x^2+ax-4

Assume that the roots are x1 and x2

then:

x1^2+ax1-4=0 ....(1)

x2^2+ax2-4=0 ....(2)

add (1) and (2)

==> x1^2+x2^2+ax1+ax2-8=0

==> x1^2+x2^2= 8-a(x1+x2)

but (x1+x2)= -a

==> x1^2+x2^2=8+a^2......(3)

Now we need to calculate the sum of the cubes of the roots:

that means we need x1^3+x2^3

Let us multiply equation (1) with x1 and equation (2) with x2

==> x1^3+ax1^2-4x1=0 .......(4)

==> x2^3+ax2^2-4x2=0 ........(5)

Now let us add (4) and (5)

==> x1^3+ax1^2-4x1+x2^3+ax2^2-4x2=0

==> x1^3+x2^3 = 4x1+4x2-ax1^2-ax2^2

factor the right side:

==> x1^3+x2^3 = 4(x1+x2)-a(x1^2+x2^2)

But from equation (3) we have x1^2+x2^2=8+a^2

==> x1^3+X2^3 = 4(-a)-a(8+a^2)

                        = -4a-8a-a^3

                        = -12-a^3

Then  x1^3+x2^3= -12-a^3

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