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Hi, I don't get what to do here I tried using vietas theorem, but I still can't get it. 55 points

Hi, I don't get what to do here I tried using vietas theorem, but I still can't get-example-1
Hi, I don't get what to do here I tried using vietas theorem, but I still can't get-example-1
Hi, I don't get what to do here I tried using vietas theorem, but I still can't get-example-2
Hi, I don't get what to do here I tried using vietas theorem, but I still can't get-example-3

1 Answer

5 votes

Answer:

x^3 - 6x^2 + 12x - 8 = 0

Explanation:

Let the roots of the equation x^3 - 8x + 3 = 0 be a, b, and c. Then, we know that the squares of the roots are a^2, b^2, and c^2.

By Vieta's formulas, we know that the sum of the roots of x^3 - 8x + 3 = 0 is 0, since there is no x^2 term:

a + b + c = 0

We also know that the product of the roots is 3/1 = 3:

abc = 3

Using the fact that the squares of the roots are a^2, b^2, and c^2, we can write:

a^2 + b^2 + c^2 = (a + b + c)^2 - 2(ab + ac + bc)

Since a + b + c = 0, this simplifies to:

a^2 + b^2 + c^2 = -2(ab + ac + bc)

Now, let's try to find a cubic equation with roots a^2, b^2, and c^2. We can start with:

(x - a^2)(x - b^2)(x - c^2) = 0

Expanding the left side, we get:

x^3 - (a^2 + b^2 + c^2)x^2 + (a^2b^2 + b^2c^2 + c^2a^2)x - a^2b^2c^2 = 0

Using the identity we found earlier for a^2 + b^2 + c^2, we can simplify this to:

x^3 + 2(ab + ac + bc)x - 3a^2b^2c^2 = 0

Now, we can substitute in the values we know for ab, ac, and bc:

ab = (a + b + c)(ab + ac + bc) - (a^2b + ab^2 + c^2a)

ac = (a + b + c)(ab + ac + bc) - (a^2c + ac^2 + b^2a)

bc = (a + b + c)(ab + ac + bc) - (b^2c + bc^2 + a^2b)

Plugging in these values and simplifying, we get:

x^3 - 6x^2 + 12x - 8 = 0

Therefore, the cubic equation we're looking for is:

x^3 - 6x^2 + 12x - 8 = 0

And the roots of this equation are the squares of the roots of x^3 - 8x + 3 = 0.

User Wildmonkey
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