Question

Pls help with 29..tank you

Answer 1

Step-by-step explanation:

If s and s^2 are the roots of the equation then

a·(x - s)·(x - s^2) = 0

a·x^2 + a·(-s - s^2)·x + a·s^3 = 0

In the equation

a·x^2 + b·x + c = 0

a = a ; b = a·(-s - s^2) ; c = a·s^3

Let's proof the equation

c·(a - b)^3 = a·(c - b)^3

a·s^3·(a - a·(-s - s^2))^3 = a·(a·s^3 - a·(-s - s^2))^3

a·s^3·(a·(1 + s + s^2))^3 = a·(a·s·(s^2 + 1 + s))^3

a^4·s^3·(1 + s + s^2)^3 = a^4·s^3·(1 + s + s^2)^3

qed.

Answer 2

Answer:

See below

Step-by-step explanation:

Let the  2 roots be  A and A^2.

Then A^3 = c/a and A + A^2 = -b/a.

Using the identity a^3 b^3 = (a + b)^3 - 3ab^2 - 3a^2b:-

A^3 + (A^2)^3 =  ( A + A^2)^3 - 3 A.A^4 - 3 A^2. A^2

= (A + A^2)^3 - 3A*3( A + A^2)

Substituting:-

c/a + c^2/a^2 = (-b/a)^3 - 3 (c/a)(-b/a)

Multiply through by a^3:-

a^2c + ac^2 = -b^3 + 3abc

Factoring:-

ac(a + c) = 3abc - b^3

This is not the formula required in the question but let's see if the formula in the question reduces to this. If it does we have completed the proof.

a(c - b)^3 = a(c^3 - 3c^2b + 3cb^2 - b^3) = ac^3 - 3ac^2b + 3acb^2 - ab^3

c(a - b)^3 = c(a^3 - 3a^2b + 3ab^2 - b^3) = ca^3 - 3a^2bc + 3acb^2 - cb^3.

These are equal so we have

ca^3 - 3a^2bc + 3acb^2 - cb^3 = ac^3 - 3abc^2 + 3acb^2 - ab^3

The 3acb^2 cancel out so we have:-

a^3c - ac^3 =  3a^2bc - 3abc^2 + b^3c - ab^3

ac(a^2 - c^2) = 3abc( a - c) + b^3(c - a)

ac(a + c)(a -c) = 3abc(a - c) - b^3 (a - c)

Divide through by (a - c):-

ac(a + c) = 3abc - b^3 , which is the result we got earlier.

This completes the proof.